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0:18
This video is about working with rational expressions. A rational expression is a fraction, usually with variables in it, something like x plus two over x squared minus three is a rational expression. In this video, we'll practice adding, subtracting, multiplying and dividing rational expressions, and simplifying them to lowest terms. We'll start with simplifying to lowest terms. Recall, if you have a fraction with just numbers in it something like 21
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over 45, we can reduce it to lowest terms by factoring the numerator and factoring the denominator and then canceling common factors. So in this example, the three is cancel, and our fraction reduces to seven over 15. If we want to reduce rational expression with variables and add to lowest terms, we proceed the same way. First, we'll factor the numerator, that's three times x plus two, and then factor the denominator. In this case of factors 2x
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plus two times x plus two, we could also write that as x plus two squared. Now we cancel the common factors. And we're left with three over x plus two. Definitely a simpler way of writing that rational expression. Now next, let's practice multiplying and dividing. Recall that if we multiply two fractions with just numbers in them, we simply multiply the numerators and multiply the denominators. So in this case, we would get four times two over three
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times five or 8/15. If we want to divide two fractions, like in the second example, then we can rewrite it as multiplying by the reciprocal of the fraction on the denominator. So here, we get four fifths times three halves, and that gives us 12 tenths. But actually, we could reduce that fraction to six fifths, we use the same rules when we compute the product or quotient of two rational expressions with variables on them. Here, we're trying
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to divide to rational expressions. So instead, we can multiply by the reciprocal, I call this flipping and multiplying. And now we just multiply the numerators. And multiply the denominators. It might be tempting at this point to multiply out to distribute out the numerator and the denominator. But actually, it's better to leave it in this factored form and factored even more completely. That way, we'll be able to reduce the rational expression
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to cancel the common factors. So let's factor even more the x squared plus x factors as x times x plus one, and x squared minus 16. And that's a difference of two squares, that's x plus four times x minus four, the denominator is already fully factored, so we'll just copy it over. And now we can cancel common factors here and here, and we're left with x times x minus four. This is our final answer. Adding and subtracting fractions is a little more
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complicated because we first have to find a common denominator. A common denominator is an expression that both denominators divided into, it's usually best of the long run to use the least common denominator, which is the smallest expression that both denominators divided into. In this example, if we just want a common denominator, we could use six times 15, which is 90 because both six and 15 divided evenly into 90. But if we want the least common denominator,
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the best way to do that is to factor the two denominators. So six is two times 315 is three times five, and then put together only the factors we need for both six and 15 to divide our numbers, so if we just use two times three times five, which is 30. We know that two times three will divide it and three times five will also divide it and we won't be able to get a denominator any smaller because we need the factor of two three and five, in
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order to ensure both these numbers divided, once we have our least common denominator, we can rewrite each of our fractions in terms of that denominator. So seven, six, I need to get a 30 in the denominator, so I'm going to multiply that by five over five, and multiply by the factors that are missing from the current denominator in order to get my least common denominator of 30. For the second fraction, for 1515 times two is 30. So I'm going to
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multiply by two over two, I can rewrite this as 3530 s minus 830 s. And now that I have a common denominator, I can just subtract my two numerators. And I get 27/30. If I factor, I can reduce this to three squared over two times five, which is nine tenths. The process for finding the sum of two rational expressions with variables in them follows the exact same process. First, we have to find the least common denominator, I'll do that by factoring
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the two denominators. So 2x plus two factors as two times x plus 1x squared minus one, that's a difference of two squares. So that's x plus one times x minus one. Now for the least common denominator, I'm going to take all the factors, I need to get an expression that each of these divides into, so I need the factor two, I need the factor x plus one, and I need the factor x minus one, I don't have to repeat the factor x plus one I just
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need to have at one time. And so I will get my least common denominator two times x plus one times x minus one, I'm not going to bother multiplying this out, it's actually better to leave it in factored form to help me simplify later. Now I can rewrite each of my two rational expressions by multiplying by whatever's missing from the denominator in terms of the least common denominator. So what I mean is, I can rewrite three over 2x plus two, I'll write
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the 2x plus two is two times x plus one, I'll write it in factored form. And then I noticed that compared to the least common denominator, I'm missing the factor of x minus one. So I multiply the numerator and the denominator by x minus one I just needed in the denominator, but I can't get away with just multiplying by the denominator without changing my expression, I have to multiply by it on the numerator and the denominator. So I'm just multiplying
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by one and a fancy form and not changing the value here. So now I do the same thing for the second rational expression. I'll write the denominator in factored form to make it easier to see what's missing from the denominator. What's missing in this denominator, compared to my least common denominator is just the factor two. So I multiply the numerator and the denominator by two. Now I can rewrite everything. So the first rational expression
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becomes three times x minus one over two times x plus 1x minus one, and the second one becomes five times two over two times x plus 1x minus one, notice that I now have a common denominator. So I can just add together my numerators. So I get three times x minus one plus 10 over two times x plus 1x minus one. I'd like to simplify this. And the best way to do that is to leave the denominator in factored form. But I do have to multiply out the numerator
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so that I can add things together. So I get 3x minus three plus 10 over two times x plus 1x minus one, or 3x plus seven over two times x plus 1x minus one. Now 3x plus seven doesn't factor. And there's therefore no factors that I can cancel out. So this is already reduced. As much as it can be. This is my final answer. In this video, we saw how to simplify rational expressions to lowest terms by factoring and
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canceling common factors. We also saw how to multiply rational expressions by multiplying the numerator and multiplying the denominator had divide rational expressions by flipping and multiplying, and how to add and subtract rational expressions by writing them in terms of the least common denominator. This video is about the difference quotient and the average rate of change. These are topics that are related to the concept of derivative and calculus.
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For function y equals f of x, like the function whose graph right here, a secant line is a lie. that stretches between two points on the graph of the function. I'm going to label this x value has a, and this x value as B. So this point here on the graph is going to have an x value of a, and a y value given by f of a, the second point will have x value B and Y value, f of b. Now the average rate of change for a function on the interval from
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a to b can be defined as the slope have the secret line between the two points A, F of A and B, F of B. In symbols, that slope m is the rise over the run, or the change in y over the change in x, which is the difference in Y coordinates f of b minus F of A over the difference in x coordinates b minus a. So this is the average rate of change. To put this in context, if for example, f of x represents the height of a tree, and x represents
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time in years, then f of b minus f of a represents a difference in height, or the amount the tree grows. And B minus A represents a difference in in years, so a time period. So this average rate of change is the amount the tree grows in a certain time period. For example, if it grows 10 inches in two years, that would be 10 inches per two years, or five inches per year would be its average rate of change its average rate of growth, let's compute
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the average rate of change for the function f of x equals square root of x on the interval from one to four. So the average rate of change is f of four minus f of one over four minus one, well, f of four is the square root of four of one's the square root of one. So that's going to be two minus one over three or 1/3. Instead of calling these two locations on the x axis, a and b, this time, I'm going to call the first location, just x and the second location, x plus h.
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The idea is that h represents the horizontal distance between these two locations on the x axis. In this notation, if I want to label this point on the graph of y equals f of x, it'll have an x coordinate of x and a y coordinate of f of x. The second point will have an x coordinate of x plus h and a y coordinate of f of x plus h. a difference quotient is simply the average rate of change using this x x plus h notation. So a different version
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represents the average rate of change of a function, f of x on the interval from x to x plus h. Equivalent way, the difference quotient represents the slope of the secant line for the graph of y equals f of x between the points with coordinates x, f of x, and x plus h, f of x plus h. Let's work out a formula for the difference quotient. Remember that the formula for the average rate of change could be written as f of b minus F of A over B minus
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A, where a and b are the two locations on the x axis. But now I'm calling instead of a I'm using x instead of B, I'm using x plus h. So I can rewrite this average rate of change as f of x plus h minus f of x over x plus h minus x. That simplifies a little bit on the denominator. Because x plus h minus x, I can cancel the access and I get the difference quotient formula f of x plus h minus f of x over h. The quantity h on denominator It
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looks like a single entity, but it still represents a difference in x values. Let's find and simplify the difference quotient for this function given first or write down the general formula for the difference quotient. That's an F of x plus h minus f of x over h. I'm going to compute f of x plus h first, I do this by shoving in x plus h, everywhere I see an x in the formula for the function. So that's going to give me two times x plus h squared
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minus x plus h plus three. Notice how I use parentheses here. That's important, because I need to make sure I shove in the entire x plus h for x. So the entire x plus h needs to be subtracted, not just the x part, so the parentheses are mandatory. Similarly, the parentheses here signal that the entire x plus h is squared as it needs to be, I'm going to go ahead and simplify a bit right now, I can multiply out the x plus h squared,
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I can go ahead and distribute the negative sign. So if I multiply out, I'm gonna get x squared plus XH plus h x plus h squared. Now I can distribute the two to get to x squared plus 2x H plus two h x plus two h squared minus x minus h plus three, these two terms are actually the same, I can add them up to get 4x H. And I think that's as simple as I can get that part. Now, I'm going to write out F of x plus h minus f of x. So that's
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going to be this thing right here, minus f of x. Again, I need to put the F of X formula in parentheses to make sure I subtract the whole thing. I'll distribute the negative. And now I noticed that a bunch of things cancel out. So the 2x squared and the minus 2x squared add to zero, the minus x and the x add to zero, and the three and the minus three, add to zero. So I'm left with 4x H plus two h squared minus h. Finally, I'll write out the whole difference quotient by dividing everything
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by H. I can simplify this further, because notice that there's an H in every single term of the numerator. If I factor out this H, H times 4x, plus two h minus one divided by h, these two H's cancel, and I'm left with a difference quotient of 4x plus two h minus one. This difference quotient will become important in calculus, when we calculate the difference quotient for smaller and smaller values of h, eventually letting h go to zero,
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and ending up with an expression that has no H's in it, and represents the derivative or slope of the function itself. In this video, we use the formula f of b minus F of A over B minus A to calculate an average rate of change, and the related formula f of x plus h minus f of x over h to calculate and simplify a difference quotient. This video will introduce the idea of limits through some graphs and examples. When I worked in San Francisco,
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I used to eat at Julia sushi and salad bar that had an interesting price structure, the food there cost $10 a pound. But if you happen to load up your plate to exactly one pound, you got a free lunch. So let's let y equals f of x represent the price of your lunch in dollars as a function of its weight x in pounds. And I want to write an equation to describe f of x as a piecewise defined function. It makes sense to use a piecewise defined function.
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Because there are two situations, the weight x could be exactly one pound, or it could be different from one pound. If the weight is exactly one pound, then f of x The price is zero. But if x is different from a pound, then the price is 10 times x as a graph, my function is going to follow the line y equals 10x. But when x is exactly one, my function is going to have a value of zero and not 10. The open circle here represents a hole or
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a place where a point is missing on the graph. Now when x is near one, but not equal to one, then the F of X values that is the y values are very close to 10. In the language of limits, we say that the limit as x approaches one of F have x is equal to Ted. More informally, we can write, as x approaches one, f of x approaches 10. Notice that the value of f at one is actually equal to zero, not 10. And so the limit of f of x as x goes to one,
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and the value of f at one, are not equal. This illustrates the important fact that the limit here as x goes to one doesn't care about the value of f, at one, it only cares about the value of f, when x is near one, in general, for any function f of x, and for real numbers, a and L, the limit as x goes to a of f of x equals L means that f of x gets arbitrarily close to L, as x gets arbitrarily close to a, in other words, as x heads towards a, f
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of x heads towards L. Let me draw this as a picture. In this picture, I can say that the limit as x goes to a of f of x is L. Because by taking a sufficiently small interval around a of x values, I can guarantee that my y values, my f of x values lie in an arbitrarily small interval around L. Now the limit doesn't care about s value at exactly a. So if I change the function, if I change the functions value at a, or even leave the function undefined
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at a, the limit is still L. But the limit does care about what happens for x values on both sides of a. If, for example, the function didn't even exist for x values bigger than a, then we could no longer say that the limit, as x approached a was out, the limit would not exist. In other words, the limit of f of x is L, only if the y values are approaching now, as x approaches a from both the left and the right. For the function g of x graph below, the limit as
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x goes to two of g of x does not exist. Because the y values approach one as the x values approach two from the left and the y values approach three, as the x values approach two from the right. Although the limit doesn't exist, we can say that the left sided limit exists. And we write this as limit as x goes to two from the left of g of x is equal to one. The superscript minus sign here means that x is going to two from the left side.
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In other words, the x values are less than two and approaching two. Similarly, we can talk about right sided limits. In this example, the limit as x goes to two from the right side of g of x is three. And here, the superscript plus side means that we're approaching two from the right side. In other words, our x values are greater than to and approaching to. In general, the limit as x goes to a minus of f of x equals L means that f of x approaches
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L, as x approaches a from the left, and the limit as x goes to a plus of f of x equals our means that f of x approaches our as x approaches a from the right limits from the left or from the right are also called one sided limits. In this last example, let's look at the behavior of y equals h of x graph below when x is near negative two. As x approaches negative two from the right, our Y values are getting arbitrarily large, larger than
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any real number we might choose. We can write this in terms of limits by saying the limit as x goes to negative two from the right of h of x is equal to infinity. As x approaches negative two from the left, the y values are getting smaller. In fact, they go below any negative real number we might choose. In terms of limits, we can say that the limit as x goes to negative two from the left of h of x is negative infinity. In this example, the
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limit of h of x as x goes to negative two, not specifying from the left or the right means we have to approach negative two from both sides. And so this limit does not exist, because the limits from the left and the limit from the right are heading in opposite directions. I want to mention that some people say that the limits from the right and the limits from the left also do not exist. Because the functions don't approach any finite number. I prefer
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to say that these limits do not exist as a finite number. But they do exist as infinity and negative infinity. This video introduced the idea of limits, and one sided limits and infinite limits. This video gives some examples of when limits fail to exist. For this function, f of x graph below, let's look at the behavior of f of x in terms of limits, as x approaches negative one, one, and two. Let's start with x approaching negative one. When x approaches
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negative one from the last, the y value is going to be approaching about one half. When x approaches negative one from the right, the y values sandvi, approaching one. So when x approaches negative one, and we don't specify from either the left or the right, we can only say that the limit does not exist, because these two limits from the left and right are not equal. Now let's look at the limit as x is approaching one. This time, we approach from the left,
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we get a limiting y value of two, we approach from the right, the y values are going towards two. So both of these left and right limits are equal to and therefore the limit as x goes to one of f of x equals two. That's true, even though f of one f of one itself does not exist. The limit doesn't care what happens at exactly x equals one just what happens when x is near one. Finally, let's look at the limit as x goes to two. So here on the left side, the limit
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is going to negative infinity. And on the right side, it's negative infinity. So we can say the limit as x goes to two is negative infinity. Or we can also say that the limit as x goes to two does not exist. This is the correct answer. This is a better answer because it carries more information. What values of x does a limit of f of x fail to exist? Well, let's see. Negative one and two are the only two values. Let's talk about the ways that
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limits can fail to exist, we've seen at least a couple different ways. So we've seen examples where the limit from the left is not equal to the limit from the right. Here's our number a where we're calculating the limit at. So that's one example we've seen. We've also seen examples where they're vertical asymptotes. There's a vertical asymptote here at a, that limit fails to exist because of the unbounded behavior because the y values are going off
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to infinity. There's one other way that limits can fail to exist that comes up, sometimes not quite as frequently. And that's wild behavior. Not a technical term, just to descriptive term. Let's look at an example that has this wild behavior, forcing a limit not to exist. And the one of the most classic examples as the limit as x goes to zero of sine pi over x or sometimes you'll see sine one over x. If you graph this on your graphing calculator
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and zoom in near x equals zero, you're gonna see something that looks roughly like this. It just keeps oscillating up and down and up and down, as x goes towards zero As x goes towards zero, pi over x is getting bigger and bigger. And you're going to go through these oscillations between one and negative one for the faster and faster. From the other side, when x is negative, you'll see a similar kind of behavior just oscillating faster and faster. As x goes to zero,
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here, this top values up here at one, and the bottom value of these are all supposed to hit it negative one. Now, when you try to decide what the limit is, as x goes to zero, well, the y values are going through all possible real numbers and between negative one and one infinitely often as x goes to zero, so there's no single number that the limit can settle at. And so the limit as x goes to zero, of sine pi over x does not exist.
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In this video, we saw three types of examples when limits fail to exist, they can fail to exist because the one sided limits on the left and the right are not equal. Or they can fail to exist because of vertical asymptotes. Also, limits can fail to exist when there's wild behavior. And the function fails to settle down at any single value. This video is about limit laws, rules for finding the limits of sums, differences, products, and quotients
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and functions. Let's start with an example. Suppose that the limit as x goes to seven of f of x is 30. And the limit as x goes to seven of g of x is two. What's the limit as x goes to seven of f of x divided by negative three times g of x. Wilson's f of x is heading towards 30. And g of x is heading towards two, it makes sense that the quotient should head towards 30 divided by negative three times two, or negative five.
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in calculating this limit, by plugging in numbers for the components, we were actually using the limit loss, which are now state. Suppose that c is a constant, just some number, and that the limits as x goes to a of f of x and g of x exist as finite numbers that is not as limits that are infinity or negative infinity. Then the limit of the sum f of x plus g of x is equal to the limit of f of x plus the limit of g of x. In other words,
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the limit of the sum is the sum of the limits. Similarly, the limit of the difference is the difference of the limits. The limit of C times f of x is just c times the limit of f of x. And the limit of the product is the product of the limits. The limit of the quotient is the quotient of the limits, provided that the limit of g of x is not equal to zero. Since we can't divide by zero. Let's use these limit laws in an example, to find the limit
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as x goes to two of x squared plus 3x plus six divided by x plus nine. Well, the limit rule about quotients allows us to rewrite this limit of a quotient as a quotient of limits, provided that the limits of the numerator and denominator exist, and that the limit of the denominator is not zero. But we'll see in a moment that these conditions do in fact hold. Next, we can use the limit rule about sums to rewrite the numerator as a sum
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of limits. And we can rewrite the limit in the denominator also as a sum of limits. Next, we can use the limit rule about products to rewrite the limit of x squared as the square of the limit. That's because x squared is really x times x. And so the limit of x squared is really the limit of x times x, which by the product rule is a limit of x times the limit of x, which is the limit of x quantity squared. Going back to the original problem,
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we can now use the limit rule about multiplying by a constant to rewrite the limit of three times x as three times the limit of x. And I'll just carry the rest forward. Now that we've got things broken down into bite sized pieces, we can start evaluating some limits. Notice that the limit as x goes to two of x is just two, because as x heads towards 2x, heads towards two, so we can replace all these limits of access by just the number
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two. So we get two squared plus three times two. Now notice that the limit as x goes to two of six, well, six doesn't have any x's in it. So as x two heads towards to six days at six, and the limit here is just six. So in my original problem, I can replace the set limit of six with six. And on the denominator here, I get to plus nine. And after a little arithmetic, this simplifies to 16/11. Notice that we could have gotten this answer a lot
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faster, just by substituting in the value of two into our original expression. And in fact, that's the beauty of the lemon laws, they allow us to evaluate limits of rational functions, just by plugging in the number that x is going towards, as long as plugging in that number doesn't make the denominator zero. It's not nearly so simple. When plugging in the value does make the done at zero. And in the future, we'll build up a bunch of algebraic
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techniques for handling the situation. In this video, we've talked about the limit loss. It's important to note, these limit laws only apply if the limits of the component functions actually exist as finite numbers. If the limit of one or both of the component functions don't exist, then a limit rules just simply don't apply. And instead, we have to use other techniques to try to evaluate the limit of a sum, difference product or quotient. This video is about the squeeze
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theorem, which is another method for finding limits. Let's start with an example. Suppose we have a function g of x. We don't know much about it. But we do know that for x value is near one, g of x is greater than or equal to four times the square root of x drawn here in red, and less than or equal to 3x squared minus 4x plus five, drawn here in blue. So on the picture, G has to lie between the red and the blue curves for x values near one,
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it could look something like this. What can we say about the limit of g of x as x goes to one? Well, if you notice, the red curve and the blue curve have the exact same limit of four as x goes to one. And since the green curve is squeezed in between the red and the blue curve, its limit must also be for this example as a special case of the squeeze theorem. Now, let's say the squeeze theorem in general, suppose that we have three functions, f of
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x g of x and h of x. And let's suppose that f of x is less than or equal to g of x, which is less than or equal to h of x, at least for x values near some number A. this inequality doesn't necessarily have to hold for x equal to a, because we're going to be talking about limits. And limits don't care what happens when x is exactly a just when x is near a. Let's suppose also, that like in the previous example, f of x and h of x have the exact
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same limit as x approaches a. So we're going to suppose that the limit as x goes to a of f of x is equal to the limit as x goes to a of h of x. And we'll call this limit. Now. The picture looks a lot like the previous example. Again, it doesn't matter exactly what happens at x equals a, for example, g of x could have a hole there. And its value could be for example, way up here. Since g of x is trapped here in between f of x and
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h of x, which both have the same limit l at a, we can conclude that the limit as x goes to a of g of x is equal to l also. And that's the squeeze theorem, also known as the pinching theorem, and the sandwich theorem, three very descriptive names that capture the idea of geovax being trapped here, in between lower and upper bounds. Now let's use the squeeze there. To find the limit as x goes to zero of x squared sine one over x. Now, you might
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remember that sine one over x by itself has this crazy oscillating behavior. In fact, the limit as x goes to zero of sine one over x does not exist, because the function never settles down to a single finite value. At first glance, you might think the limit of x squared sine of one over x also wouldn't exist. In fact, it's tempting to try to use the product rule and say that the limit of the product is the product of the limits.
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But in fact, the product rule only applies when the component limits both exist. And since the second limit doesn't exist, the product rule tells us absolutely nothing about whether the limit that we're interested in exists or doesn't. So we can't use the product rule. But it turns out, we can use the squeeze theorem. Now this example is a little trickier than the first example, because in the first example, we were told what the upper and lower
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bounding functions should be. And in this example, we have to come up with. But if we look at a graph of x squared sine one over x, we can see that it does seem to be trapped in an envelope here. Let's use algebra to see what those two bounding functions might be. Now we know that sine of one over x is always between one and negative one, just because sine of anything has between one and negative one. And if we multiply this whole inequality by x squared, we get minus
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x squared is less than or equal to x squared, sine one over x, which is less than or equal to x squared. Notice that x squared is always positive, so we don't have to worry about flipping any of the inequality signs when we multiply by this positive number. So x squared and minus x squared are good bounding functions. And if we notice that the limit as x goes to zero of x squared is zero, and the limit as x goes to zero of negative x
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squared is also zero, we can conclude by the squeeze theorem, that the limit as x goes to zero of x squared, sine of one over x is also zero, because it squeezed in between these two functions with the same limit. the squeeze theorem is a great trick for evaluating limits when you happen to have a function that you're interested in, trapped in between two other functions with the same limit. The example in this video is a classic example,
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where we have a crazy oscillating trig function, multiplied by a power of x, that x is one of our bounding functions. In this video, we'll compute a bunch of limits using algebraic tricks. All these limits are of the zero over zero indeterminate form kind. Recall that this means that the limits are of the form the limit of f of x over g of x where the limit of f of x equals zero and the limit of g of x also equals zero. For a first example,
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let's look at the limit as x goes to one of x cubed minus one over x squared minus one. Notice that the numerator and the denominator are both going to zero as x goes to one. To calculate this limit, we want to simplify this expression. And one way to simplify it is to factor it. So let's rewrite this as the limit as x goes to one of x minus one times x squared plus x plus one. We're factoring the numerator here is a difference of cubes.
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We can also factor the denominator as a difference of squares, x minus one times x plus one. Now, as long as x is not equal to one, we can cancel out these two factors of x minus one. And so the limit of this expression is just the same as the limit of this expression. Now we can just plug in one, because plugging in one gives us a numerator of three and a denominator of two. And that way, we've evaluated our limits. Next, let's look at the limit
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of five minus z quantity squared minus 25. All divided by z. Once again, when we plug in x equals zero in the numerator, we get zero and then nominator we also get zero. This time instead of factoring, the trick is going to be to multiply out. So let's rewrite this limit as the limit as t goes to zero of 25 minus 10 z plus z squared minus 25. I just distributed to get this expression divided by z. Since 25 minus 25 is zero, I just have the limit as t goes to zero of negative 10
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z plus c squared over z. Now what? Well, I could factor out the Z here. One z is not zero, I can cancel here. So my original limit is the same as this limit. Plugging in z equals zero, I just get negative 10 as my answer. This third example is also a situation where the numerator is going to zero, and the denominator is also going to zero. In this case, I'm going to try to simplify the expression by adding together the fractions and the numerator. So I'll need a common denominator,
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which is r plus three, times three. So rewriting, I get the limit of one over r plus three, I multiply that by three over three, in order to get the appropriate common denominator, minus 1/3, which I have to multiply by R plus three over r plus three, all that over are continuing to rewrite, I have in the numerator, adding together these fractions mature, have the denominator of R plus three times three, I get three minus quantity r plus three. And
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then this entire fraction is still divided by our distributing the negative sign, I have three minus r minus three divided by R plus three times three, all divided by R. three minus three is zero. So I can rewrite this as negative r over r plus three, times three. And now instead of dividing by r, which is r of r one, I can multiply by the reciprocal, one over R. R divided by R is one. So this expression simplifies to negative one, over
48:11
r plus three, times three. So finally, I'm in a good position, because now I can just go ahead and let our go to zero. And by plugging in R equals zero, I have a limit of negative one over zero plus three times three, or negative one, nine. This example is a little tricky, because involve square roots can be hard to deal with. Well, there's one nice trick for dealing with square roots that works here, which is the conjugate. So I'm going to take
48:46
the expression that we're given, and multiply by the conjugate of the numerator, in this case, because the numerator is the place where the square root is by the conjugate of A minus B, I just mean a plus b conjugate of A plus B is a minus b. Well, of course, if I multiply on the numerator, by something, I also have to multiply the denominator by the same thing, so that I won't alter the value of the expression. So this limit here is equal to the limit of
49:24
the expression down here that looks more complicated, than in a moment, if we're lucky, things will clear up and become simpler. So Multiplying the numerators across, I get the square root of x plus three squared. I get plus two times the square root of x plus three minus two times the square root of x plus three minus four. On the denominator, I get x times the square root of x plus three plus 2x minus the square root of x plus three minus two. Let's see what simplifies here. So the
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square root of x plus three squared is just x plus three. And I see that this expression and this expression are opposites, so they subtract to zero here. So on the numerator, I just have x plus three, and then I still have the minus four on the denominator, nominator looks a little messy. I'll just copy it over for now. Okay, so on my numerator, I'm getting x minus one. Notice that when x goes to one, that numerator is still going to zero. And
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in fact, as I let x go to one, that denominator, if I plug in here, everything is gonna also cancel out to zero. So I still got a zero over zero indeterminate form. But maybe I can use one of the previous tricks of factoring. Because look at here, if I factor an X out of these two expressions as a square root of x plus three out of these two expressions, and I could maybe factor a two out of these two, factoring by grouping, let's try that.
51:28
So we have x squared of x plus three times x minus one from these two expressions, and then I have a plus two times x minus one. From these two expressions, this is looking promising. So now I've got an x minus one on the top. And if I factor out the x minus one from each of these two expressions, I'm going to have an x minus one on the bottom times the square root of x plus three, plus two. Now, for x values near one, but not equal
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to one, I can cancel those. And my limit simplifies to just one over square root of x plus three plus two, plugging in one, I get a one on the numerator, and square root of four, which is two plus two, four on the denominator, and we have calculated this limit. Last example here, another zero over zero indeterminate form. Anytime I see an absolute value, I'm going to want to take cases, because the absolute value of X plus five naturally follows in
53:05
the cases, if X plus five is greater than zero, in other words, x is greater than negative five, then the absolute value of this positive number is just itself. On the other hand, if X plus five is less than zero, in other words, x is less than negative five, than the absolute value of a negative number is its opposite. And we make the expression x plus five, turn it into its opposite by putting a negative sign in
53:37
front. Now let's look at one sided limits. As x goes to negative five from the left, we have a situation where x is less than negative five, this situation right here. And so we can rewrite the absolute value by taking the negative of the expression. We still have a zero over zero and determinant form. But if we use the old factoring trick, factor out a two cancel, we got the limit of two over negative one, which is just negative two.
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We do the same exercise on the right side. When we're on the right, then x is greater than negative five. So we're in this situation here, where we can just replace the absolute value with the stuff inside and again factoring the numerator and canceling the X plus five We just get a limit of two. So we have a left limit, and a right limit that are different. And so in this example, the limit does not exist. So we've seen a five different kinds of limits all of the
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zero or zero indeterminate form. And we've used five different methods to evaluate them. For the first example, we use factoring. For the second example, do the opposite of factoring, we multiplied out. For the third example, we added together our rational expressions to simplify things. The next example uses the old multiply by the conjugate trick. And the last example, we used cases and looked at one sided limits.
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The limit law about quotients tells us that the limit of the quotient is the question of the limit, provided that the limits of the component functions actually exist, and that the limit of the function on the denominator is not equal to zero. But what happens if the limit of the function on the denominator is equal to zero? This video will begin to answer that question. In fact, there are two different situations we'll want to consider.
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It could be that even though the limit on the denominator is equal to zero, the limit on the numerator exists and is not equal to zero. Or it could happen that both limits are zero. We'll focus on the first situation, first starting with an example. And we'll look at the second situation later on. In this example, the limit of the numerator, negative 4x is just negative 12, which we can see by plugging in three for x. But the limit as x goes to three of the denominator
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is zero. So we're exactly in one of these situations where the numerator goes to a finite nonzero number, but the denominator goes to zero. Let's see what happens as we approach three from the left first. As we approach three from the left, x is going through numbers that are slightly less than three numbers like 2.9 2.99 2.999, and so on. If we plug in those numbers into the expression, here on our calculator, we didn't get answers of
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116 1196, and 11,996. If even without a calculator, we could approximate these answers pretty closely by just thinking about the fact that since x is very close to three, the numerator is about negative four times three, so about negative 12. The denominator 2.9 minus three is negative 0.1. That quotient of two negative numbers gives us a positive value of 120. Similarly, we could approximate the value when x is 2.9 times as almost negative 12
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divided by 0.01, which is 1200. And approximate the third value as 12,000. Either way, we do it exact answers on our calculator or approximations in our head when noticing that these values are positive numbers that are getting larger and larger as x goes towards three from the left. This makes sense. Because if we look at our expression, as x goes towards three from the left, the numerator is getting close to negative 12, which is a negative number.
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And the denominator, since x is less than three will always be a small negative number, negative over negative is a positive. And as x is getting really close to three, those denominators are getting smaller and therefore the fractions are getting bigger and bigger in magnitude. So we can conclude that our limit is positive infinity. We can make a similar argument By looking at the limit, as x goes to three from the positive side,
1:00:08
it's supposed to be an X minus three here. So now x is going through a value is slightly bigger than three 3.1 3.01 3.01. And again, we can plug directly into our calculator and figure out the answers are negative 124. Negative 1204, negative 12,004. Or we can make a similar approximating argument. This answer is approximately negative 12 over a positive point one, which is negative 120, and so on. Like before, if we consider the signs of our numerator and
1:01:01
denominator, we can see that as x goes to three, our numerator is a negative number. But our denominator is a positive number, since we're approaching three from the right where x is bigger than three, and therefore, our quotient is a negative number, it's still getting bigger and bigger in magnitude as x goes towards three, because the denominator is still getting tinier and tinier, while the numerator stays pretty close to negative
1:01:27
12. So in this case, we're getting a negative number that's bigger and bigger magnitude, so that makes a limit of negative infinity. Now, since our limit on the left is infinity, and our limit on the right is negative infinity, the only thing we can say about the limit as x goes to three is that it does not exist. Now let's look at another example. The limit as x goes to negative four of 5x, over the absolute value of x plus four, notice that the limit of the
1:02:09
numerator is just negative 20 by plugging in negative four for x, and the limit of the denominator is zero. Because we've got an absolute value in our expression, here, it's screaming out at us to look at cases. Remember that the absolute value of x plus four is going to equal just x plus four, if x plus four is positive, in other words, if x is greater than negative four, however, if x is less than negative four, then the expression
1:02:54
x plus four will be negative. So taking the absolute value has to switch at sign in order to make a negative expression positive. Alright, so that's going to come in handy when we look at the limit as x goes to negative four, from the left, and from the right. So when we approach negative four from the left, x is going to be less than negative four. So we're going to be in this situation here, where the absolute value gives us the opposite sign. That means
1:03:42
that the limit as x goes to negative four minus of this expression, the same as the limit of 5x over negative x plus four. Now reasoning as before, as x is going to negative four from the left, the numerator here is a negative number. Pretty close to negative 20. The denominator, since x is less than negative 4x plus four is negative, the negative of it is positive. So our quotient is negative. And since denominator is getting really tiny,
1:04:28
while the numerators will pretty level at at negative 20. This limit is going to be bigger and bigger magnitude a limit of negative infinity. Now let's look at the limit as x goes to negative four from the right, in this case, x is just a little bit bigger than negative four. So we're in this case where the absolute value doesn't change the expression. So we can rewrite this As 5x over x plus four, now the numerator is still gonna be a negative number, the denominator, since x is slightly
1:05:10
bigger than negative four, slightly to the right, this expression is going to be positive number. Negative or over a positive is a negative. And again, since the denominator is getting tiny, the fractions getting huge in magnitude. And so this limit is negative infinity. Now, in this case, look at what's going on, we've got a negative infinity limit on the left and a negative infinity limit on the right. So we can conclude that the limit as x goes
1:05:47
to negative four of 5x, over the absolute value of x plus four is equal to negative infinity. In fact, we can confirm that by looking at a graph. If we look, check out the graph near x equals negative four, it's gonna look something like this, with a vertical asymptote at x equals negative four, like expected. So we've seen that if the limit of f of x is equal to something that's not zero, and the limit of g of x is equal to
1:06:23
zero, then the limit of the quotient could be negative infinity as it was in the past example. It could also be infinity, or it could just not exist. supposed to say does not exist, if the one sided limits are infinity on one side, and negative infinity on the other. Now, what about this second situation that I mentioned the beginning when the limit of f of x is zero, and the limit of f of g of x is zero? What can we say about the limit
1:07:09
of the quotient in this situation? Well, in fact, in this situation, the limit of the quotient, it could exist and be any finite number, or infinity, or minus infinity, or it could not exist at all. In fact, in this sort of situation, which is called a zero, or zero indeterminate form, anything could happen. Which makes it in some ways the hardest, but in some ways, the most fun situation of all. So in another video, we'll talk about
1:07:42
techniques for dealing with 00 indeterminate forms, and how to use algebra and other simplification techniques to evaluate these these mysterious limits. So in this video, we've looked at the limits of quotients, when the limit of the denominator is zero. We've done some examples when the limit of the numerator was not zero, but the limit of the denominator was zero. And we saw that these situations corresponded to vertical asymptotes, and gave us an answer
1:08:16
for the limit of the quotient of infinity, or negative infinity, or sometimes infinity on one side and negative infinity on the other. We also hinted at fun things to come when we look at the limits in this situation when the numerator and the denominator are both heading towards zero. And when anything can happen. This video is about graphs and equations of lines. Here we're given the graph of a line, we want to find the equation, one standard
1:08:50
format for the equation of a line is y equals mx plus b. here, m represents the slope, and B represents the y intercept, the y value, where the line crosses the y axis, the slope is equal to the rise over the run. Or sometimes this is written as the change in y values over the change in x values. Or in other words, y two minus y one over x two minus x one, where x one y one and x two y two are points on the line.
1:09:31
While we could use any two points on the line, to find the slope, it's convenient to use points where the x and y coordinates are integers. That is points where the line passes through grid points. So here would be one convenient point to use. And here's another convenient point to use. The coordinates of the first point are one, two, and the next point this is let's say, five negative one. Now I can find this slope by looking at the rise over
1:10:01
the run. So as I go through a run of this distance, I go through a rise of that distance, especially gonna be a negative rise or a fall because my line is pointing down. So let's see counting off squares, this is a run of 1234 squares and a rise of 123. So negative three, so my slope is going to be negative three, over four. I got that answer by counting squares. But I could have also gotten it by looking at the difference in my y values over
1:10:36
the difference of my x values. That is, I could have done negative one minus two, that's from my difference in Y values, and divide that by my difference in x values, which is five minus one, that gives me negative three over four, as before, so my M is negative three fourths. Now I need to figure out the value of b, my y intercept, well, I could just read it off the graph, it looks like approximately 2.75. But if I want to be more
1:11:14
accurate, I can again use a point that has integer coordinates that I know it's exact coordinates. So either this point or that point, let's try this point. And I can start off with my equation y equals mx plus b, that is y equals negative three fourths x plus b. And I can plug in the point one, two, for my x and y. So that gives me two equals negative three fourths times one plus b, solving for B. Let's see that two equals negative three
1:11:51
fourths plus b. So add three fourths to both sides, that's two plus three fourths equals b. So b is eight fourths plus three fourths, which is 11 fourths, which is actually just what I eyeballed it today. So now I can write out my final equation for my line y equals negative three fourths x plus 11 fourths by plugging in for m and b. Next, let's find the equation for this horizontal line. a horizontal line has slope zero. So if we think of it
1:12:28
as y equals mx plus b, m is going to be zero. In other words, it's just y equals b, y is some constant. So if we can figure out what that that constant y value is, it looks like it's to let's see, this three, three and a half, we can just write down the equation directly, y equals 3.5. For a vertical line, like this one, it doesn't really have a slope. I mean, if you tried to do the rise over the run, there's no run. So you'd I guess you'd
1:13:04
be divided by zero and get an infinite slope. But But instead, we just think of it as an equation of the form x equals something. And in this case, x equals negative two, notice that all of the points on our line have the same x coordinate of negative two and the y coordinate can be anything. So this is how we write the equation for a vertical line. In this example, we're not shown a graph of the line, we're just get told that it goes
1:13:33
through two points. But knowing that I go through two points is enough to find the equation for the line. First, we can find the slope by taking the difference in Y values over the difference in x values. So that's negative three minus two over four minus one, which is negative five thirds. So we can use the standard equation for the line, this is called the slope intercept form. And we can plug in negative five thirds. And we can use one point, either one will do will
1:14:17
still get the same final answer. So let's use one two and plug that in to get two equals negative five thirds times one plus b. And so B is two plus five thirds, which is six thirds plus five thirds, which is 11 thirds. So our equation is y equals negative five thirds x plus 11 thirds. This is method one. method two uses a slightly different form of the equation. It's called the point slope form and it goes y minus y naught is equal
1:14:55
to m times x minus x naught where x naught Why not is a point on the line, and again is the slope. So we calculate the slope the same way, by taking a difference in Y values over a difference in x values. But then we can simply plug in any point. For example, the point one, two will work, we can plug one in for x naught and two in for Y not in this point slope form, that gives us y minus two is equal to minus five thirds x minus
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one. Notice that these two equations, while they may look different, are actually equivalent. Because if I distribute the negative five thirds, and then add the two to both sides, I get the same equation as above. So we've seen two ways of finding the equation for the line is in the slope intercept form, and using the point slope form. In this video, we saw that you can find the equation for a line if you know the slope. And you know one point, you can also find the equation
1:16:27
for the line if you know two points, because you can use the two points to get the slope and then plug in one of those points. To figure out the rest of the equation. We saw two standard forms for the equation of a line the slope intercept form y equals mx plus b, where m is the slope, and B is the y intercept. And the point slope form y minus y naught equals m times x minus x naught, where m again is the slope and x naught y naught is a point
1:17:06
on the line. This video is about rational functions and their graphs. Recall that a rational function is a function that can be written as a ratio or quotient of two polynomials. Here's an example. The simpler function, f of x equals one over x is also considered a rational function, you can think of one and x as very simple polynomials. The graph of this rational function is shown here. This graph looks different from the graph of a
1:17:42
polynomial. For one thing, its end behavior is different. The end behavior of a function is the way the graph looks, when x goes through really large positive, or really large negative numbers, we've seen that the end behavior of a polynomial always looks like one of these cases. That is why marches off to infinity or maybe negative infinity, as x gets really big or really negative. But this rational function has a different type of end behavior.
1:18:11
Notice, as x gets really big, the y values are leveling off at about a y value of three. And similarly, as x values get really negative, our graph is leveling off near the line y equals three, I'll draw that line, y equals three on my graph, that line is called a horizontal asymptote. A horizontal asymptote is a horizontal line that our graph gets closer and closer to as x goes to infinity, or as X goes to negative infinity or both. There's something
1:18:45
else that's different about this graph from a polynomial graph, look at what happens as x gets close to negative five. As we approach negative five with x values on the right, our Y values are going down towards negative infinity. And as we approach the x value of negative five from the left, our Y values are going up towards positive infinity. We say that this graph has a vertical asymptote at x equals negative five. vertical asymptote
1:19:14
is a vertical line that the graph gets closer and closer to. Finally, there's something really weird going on at x equals two, there's a little open circle there, like the value at x equals two is dug out. That's called a hole. A hole is a place along the curve of the graph where the function doesn't exist. Now that we've identified some of the features of our rational functions graph, I want to look back at the equation and see how we could
1:19:44
have predicted those features just by looking at the equation. To find horizontal asymptotes. We need to look at what our function is doing when x goes through really big positive or really big negative numbers. Looking at our equation for our function, numerator is going to be dominated by the 3x squared term when x is really big, right, because three times x squared is going to be absolutely enormous compared to this negative 12. If x is a big
1:20:12
positive or negative number in the denominator, the denominator will be dominated by the x squared term. Again, if x is a really big positive or negative number, like a million, a million squared will be much, much bigger than three times a million or negative 10. For that reason, to find the end behavior, or the horizontal asymptote, for our function, we just need to look at the terms on the numerator and the term on the denominator that have
1:20:42
the highest exponent, those are the ones that dominate the expression in size. So as x gets really big, our functions y values are going to be approximately 3x squared over x squared, which is three. That's why we have a horizontal asymptote at y equals three. Now our vertical asymptotes, those tend to occur where our denominator of our function is zero. That's because the function doesn't exist when our denominator is zero.
1:21:14
And when we get close to that place where our denominator is zero, we're going to be dividing by tiny, tiny numbers, which will make our Y values really big in magnitude. So to check where our denominators zero, let's factor our function. In fact, I'm going to go ahead and factor the numerator and the denominator. So the numerator factors, let's see, pull out the three, I get x squared minus four, factor in the denominator, that factors
1:21:40
into X plus five times x minus two, I can factor a little the numerator a little further, that's three times x minus two times x plus two over x plus 5x minus two. Now, when x is equal to negative five, my denominator will be zero, but my numerator will not be zero. That's what gives me the vertical asymptote at x equals negative five. Notice that when x equals two, the denominators zero, but the numerator is also zero. In fact, if I cancelled
1:22:18
the x minus two factor from the numerator and denominator, I get a simplified form for my function that agrees with my original function, as long as x is not equal to two. That's because when x equals two, the simplified function exists, but the original function does not it's zero over zero, it's undefined. But for every other x value, including x values near x equals two, our original function is just the same as this function. And that's why
1:22:51
our function only has a vertical asymptote at x equals negative five, not one at x equals two, because the x minus two factor is no longer in the function after simplifying, it does have a hole at x equals two, because the original function is not defined there, even though the simplified version is if we want to find the y value of our hole, we can just plug in x equals two into our simplified version of our function, that gives a y value
1:23:21
of three times two plus two over two plus seven, or 12 ninths, which simplifies to four thirds. So our whole is that to four thirds. Now that we've been through one example in detail, let's summarize our findings. We find the vertical asymptotes and the holes by looking where the denominator is zero. The holes happen where the denominator and numerator are both zero and those factors cancel out. The vertical asymptotes are all other x values where the
1:23:57
denominator is zero, we find the horizontal asymptotes by considering the highest power term on the numerator and the denominator, I'll explain this process in more detail in three examples. In the first example, if we circle the highest power terms, that simplifies to 5x over 3x squared, which is five over 3x. As x gets really big, the denominator is going to be huge. So I'm going to be dividing five by a huge, huge number, that's going
1:24:31
to be going very close to zero, and therefore we have a horizontal asymptote at y equals zero. In the second example, the highest power terms, 2x cubed over 3x cubed simplifies to two thirds. So as x gets really big, we're going to be heading towards two thirds and we have a horizontal asymptote at y equals two thirds. In the third example, the highest power terms x squared over 2x simplifies to x over two. As x gets really big, x over two
1:25:08
is getting really big. And therefore, we don't have a horizontal asymptote at all. This is going to infinity, when x gets through goes through big positive numbers, and is going to negative infinity when x goes through a big negative numbers. So in this case, the end behavior is kind of like that of a polynomial, and there's no horizontal asymptote. In general, when the degree of the numerator is smaller than the degree of the denominator,
1:25:41
we're in this first case where the denominator gets really big compared to the numerator and we go to zero. In the second case, where the degree of the numerator and the degree of the denominator equal, things cancel out, and so we get a horizontal asymptote at the y value, that's equal to the ratio of the leading coefficients. Finally, in the third case, when the degree of the numerator is bigger than the degree of the denominator,
1:26:09
then the numerator is getting really big compared to the denominator, so we end up with no horizontal asymptote. Finally, let's apply all these observations to one more example. Please pause the video and take a moment to find the vertical asymptotes, horizontal asymptotes and holes for this rational function. To find the vertical asymptotes and holes, we need to look at where the denominator is zero. In fact, it's going to be handy to factor both the numerator and
1:26:38
the denominator. Since there if there are any common factors, we might have a hole instead of a vertical asymptote. The numerator is pretty easy to factor. Let's see, that's 3x times x plus one for the denominator, first factor out an x. And then I'll factor some more using a guess and check method. I know that I'll need a 2x and an X to multiply together to the to x squared, and I'll need a three and a minus one or also minus three and a
1:27:12
one. Let's see if that works. If I multiply out 2x minus one times x plus three, that does give me back my 2x squared plus 5x minus three, so that checks out. Now, I noticed that I have a common factor of x in both the numerator and the denominator. So that's telling me I'm going to have a hole at x equals zero. In fact, I could rewrite my rational function by cancelling out that common factor, and that's equivalent, as long as x is not equal to zero. So the y value of my whole
1:27:48
is what I get when I plug zero into my simplified version, that would be three times zero plus one over two times zero minus one times zero plus three, which is three over negative three or minus one. So my whole is at zero minus one. Now all the remaining places in my denominator that make my denominator zero will get me vertical asymptotes. So I'll have a vertical asymptote, when 2x minus one times x plus three equals zero, that is, when 2x minus
1:28:25
one is zero, or x plus three is zero. In other words, when x is one half, or x equals negative three. Finally, to find my horizontal asymptotes, I just need to consider the highest power term in the numerator and the denominator. That simplifies to three over 2x, which is bottom heavy, right? When x gets really big, this expression is going to zero. And that means that we have a horizontal asymptote at y equals zero. So we found the major features
1:29:03
of our graph, the whole, the vertical asymptotes and the horizontal asymptotes. Together, this would give us a framework for what the graph of our function looks like. horizontal asymptote at y equals zero, vertical asymptotes at x equals one half, and x equals minus three at a hole at the point zero minus one. plotting a few more points, or using a graphing calculator of graphing program, we can see that our actual function will look something like this. Notice
1:29:39
that the x intercept when x is negative one corresponds to where the numerator of our rational function or reduced rational function is equal to zero. That's because a zero on the numerator that doesn't make the denominator zero makes the whole function zero. And an X intercept is where the y value of the whole function is zero. In this video, we learned how to find horizontal asymptotes of rational functions. By looking at the highest power
1:30:07
terms, we learned to find the vertical asymptotes and holes. By looking at the factored version of the functions, the holes correspond to the x values that make the numerator and denominator zero, his corresponding factors cancel. The vertical asymptotes correspond to the x values that make the denominator zero, even after factoring any any common any common factors in the numerator denominator. This video focuses on the behavior of functions and graphs, as
1:30:39
x goes through arbitrarily large positive and negative values. You may have touched on these ideas before in the past when you studied horizontal asymptotes. But in this video, we'll talk about the same ideas in the language of limits. In this first example, what happens to the function f of x drawn below as x go through larger and larger positive numbers? Well, the arrow here on the end is supposed to mean the trend continues. So as
1:31:07
x gets bigger and bigger, the values of y, that is f of x, get closer and closer to one, we can write this in the language of limits by saying the limit as x goes to infinity of f of x is equal to one. Now what happens to this function as x goes through larger and larger negative numbers, by larger and larger negative numbers, I mean numbers that are negative but are larger and larger in magnitude. So for example, negative five,
1:31:38
negative 10, negative 100, and negative a million and so on. Well, assuming this trend continues, it looks like f of x, even though it's oscillating, it's settling down at a value of two. So we say that the limit as x goes to negative infinity, of f of x equals to limits in which x goes to infinity, or negative infinity are called limits at infinity. The phrase limits at infinity should be contrasted with the phrase infinite limits. An infinite
1:32:15
limit means that the y values, or the F of X values go to infinity, or negative infinity. limits and infinity correspond to horizontal asymptotes, as drawn above, while infinite limits correspond to vertical asymptotes. The only exception to this when we don't have a horizontal or vertical asymptote is when x and f of x are both going infinite at the same time, we'll see an example of this on the next page. Let's figure out the limits
1:32:47
of infinity for these two functions, g of x and h of x. The function g of x is actually the function e to the minus x, and it has a horizontal asymptote, heading right here at y equals zero. So the limit as x goes to infinity of g of x equals zero. But as we head to the left, and x goes through larger and larger negative values, our Y values don't settle down to a particular finite value, they get arbitrarily large. And so we say
1:33:23
that the limit as x goes to minus infinity of g of x is equal to infinity. Now let's look at the graph of y equals h of x. Please pause the video for a moment and try to figure out the limits of infinity for this function. The limit as x goes to infinity of h of x is negative infinity. Because as x goes to infinity, the y values get below or more negative than any finite number. Now as X goes to negative infinity, the y values oscillate, and never
1:34:00
settle down at a particular number. So we say that the limit as x goes to minus infinity of h of x does not exist. Finally, let's look at some limits of infinity of functions without looking at their graphs first, to find the limit as x goes to infinity of one over x. Let's think about what happens to one over x. As x gets bigger and bigger through positive numbers. As x gets bigger and bigger, one over x gets smaller and smaller. So the limit
1:34:36
as x goes to infinity of one over x equals zero. To find the limit of one over x as x goes to negative infinity, let's look at what happens as X goes to negative numbers that are larger and larger in magnitude. Now one of our x goes through numbers that are negative, but they're still getting smaller and smaller. magnitude. So the limit as x goes to negative infinity of one over x is also zero. We can use similar reasoning to find the limit as
1:35:11
x goes to infinity of one over x cubed. As x goes to infinity, x cubed also goes to infinity. So one of our x cubed has to go to zero. To find the limit as x goes to infinity of one over the square root of x, notice that as x goes to infinity, the square root of x still goes to infinity. So one over the square root of x also goes to zero. In other words, both of these limits are equal to zero. Both of these examples are actually closely related,
1:35:47
because both have the form of the limit as x goes to infinity of one over x to the R, where R is a number greater than zero. In the second example, the square root of x is really x to the R, where R is one half. In fact, the limit as x goes to infinity of one over x to the R is always equal to zero. Whenever R is bigger than zero, we can even say the same thing about the limit as x goes to negative infinity of one over x to the R. As long as
1:36:25
we avoid exponents, like one half that don't make sense for negative numbers. But for other values of r, as x goes to negative infinity, x to the R is getting bigger and bigger and magnitude. And so one over x to the R is getting smaller and smaller and magnitude and heading towards zero. Notice that this is no longer true. if r is less than zero, for example, something like r equals negative two, because one over x to the minus two is really x squared.
1:37:01
And the limit as x goes to infinity of x squared is going to be infinity, not zero. In this video, we looked at examples of limits as x goes to infinity, and x goes to minus infinity. And we saw that those limits could be zero. Any other number, infinity, negative infinity, or not exist. This video gives some algebraic techniques for computing the limits at infinity of rational functions. Let's find the limit as x goes to infinity of this rational
1:37:43
function. The numerator and the denominator of this rational function are each getting arbitrarily large as x goes to infinity. One way to see this is by estimating the graphs, the graph of the numerator looks like a parabola pointing upwards. And the graph of the denominator looks like some kind of cubic. So something like this, for both of these graphs, as x goes to infinity, y also goes to infinity. So this is an infinity over infinity indeterminate
1:38:15
form. And just like the zero have over zero indeterminate forms we saw earlier, and infinity over infinity and determinant form could turn out to be absolutely anything. So we're going to use algebra to rewrite this expression in a different form that makes it easier to evaluate. Specifically, we're going to factor out the highest power of x that we can find from the numerator, and then from the denominator. In the numerator, I'm going to factor out
1:38:45
the highest power I see in the numerator, which is x squared. When I factor x squared out of 5x squared, I get five, when I factor x squared out of negative 4x, that's like dividing negative 4x by x squared, so I get negative four divided by x. You can check this works by distributing the x squared and making sure we get back to the original expression. Now the highest power of x SC, and the denominator is x cubed. So I'll factor out an x cubed
1:39:19
for each from each of those terms, I get a two minus 11 over x plus 12 over x squared. Because factoring out an x cubed is the same as dividing each term by x cubed and then writing the x cubed on the side. Now, we can rewrite again by canceling an x squared from the top and the bottom to get the limit of one over x times five minus four over x over two minus 11 over x plus 12 over x squared. Now as x goes to infinity, four over x goes
1:40:00
to zero, because I'm dividing for by larger and larger numbers. Similarly 11 over x goes to zero, and 12 over x squared goes to zero. So I end up with one over x, which is itself going to zero times something that's going to five halves. So my limit is going to be zero times five halves, which is just zero. I've actually been applying limit laws to do these last steps, which is fine, because my component limits exist as finite numbers.
1:40:34
Something that wasn't true from my original expression when I had infinity over infinity. In this example, we're asked to find the limit as x goes to negative infinity of a different rational expression. I encourage you to stop the video and try it for yourself first. In this example, the highest power of x in the numerator is x cubed, and the highest power in the denominator is also x cubed. Factoring out the x cubed from the numerator, we get
1:41:06
x cubed times three plus six over x plus 10 over x squared, plus two over x cubed. And factoring out the x cubed from the denominator, we get x cubed times two, plus one over X plus five over x cubed. Now the x cubes cancel, and all these parts go to zero. So when the dust clears here, our limit is just three halves. In this next example, the highest power in the numerator is x to the fourth, and the highest power in the denominator is
1:41:45
x squared. So we factor out the x to the fourth from the numerator and the x squared from the denominator and cancel as much as we can. A lot of these pieces are going to zero. So our limit is the same as the limit of x squared times one over negative five. As x goes to negative infinity, x squared is positive and goes towards positive infinity. multiplying by negative fifth turns it negative, but doesn't change the fact that the magnitudes are getting
1:42:22
arbitrarily large. Therefore, our final limit is negative infinity. Now let's look at the same three examples again, more informally, using a heuristic to get the same conclusions. In the first example, the term 5x squared dominates the numerator, because x squared is much larger than x when x is large. In the denominator, the highest power of x to x cubed dominates, because x cubed is much larger than x squared, or x when x is large.
1:43:00
If we ignore all the other terms in the numerator and denominator, and just focus on the important terms, which have the highest powers, then we can rewrite our limit as the limit of 5x squared over 2x cubed, which is the same as the limit as x goes to infinity of five over 2x, just by canceling Xs, which is zero as x goes to infinity. Similarly, if we just focus on the highest power terms in the numerator and denominator in the second example, we
1:43:35
get the limit of 3x cubed over 2x cubed, which simplifies to the limit of three halves, which is just three halves. In the third example, the highest power terms are x to the fourth and negative 5x squared. And we rewrite the limit using only these highest power terms and simplify, and we get the limit as x goes to negative infinity of x squared over negative five, which is negative infinity as before, for rational functions, in general, looking
1:44:13
at the highest power terms, lets you reliably predict the limits that infinity and negative infinity when the degree of the numerator is less than the degree of the denominator, then the limit as x goes to infinity, or negative infinity, is zero. As an example, one above, when the degree of the numerator is equal to the degree of the denominator, then the limit is just the quotient of the highest power terms, which is how we got three halves
1:44:43
as the limit in the second example. And finally, if the degree of the numerator is greater than the degree of the denominator, then the limit is going to be plus or minus infinity. Like it was in the third example. These shortcut rules are really handy. But it's important to also understand the technique of factoring out highest power terms. Since this technique can be used more generally. This video gave two methods for computing limits and infinity
1:45:14
of rational functions. First, there's the formal method of factoring out highest power terms and simplifying. Second, there's the informal method of looking at the degree of the numerator and the degree of the denominator, and focusing on the highest power terms. In the past, you may have heard an informal definition of continuity. Something like a function is continuous, if you can draw it without ever picking up your pencil. In this video, we'll
1:45:46
develop a more precise definition of continuity based on limits. It can be helpful to look at some examples of functions that are discontinuous, that is functions that fail to be continuous. In order to better understand what it means to be continuous. Please pause this video and try to draw graphs of at least two different functions that fail to be continuous in different ways. One common kind of discontinuity is called a jump discontinuity. A function has
1:46:18
a jump discontinuity, if its graph separates the two pieces with a jump in between them. This particular function can be described as a piecewise defined function with two linear equations, f of x equals to x when x is less than or equal to one, and f of x equals negative x plus two, when x is greater than one. Another common kind of discontinuity is called a removable discontinuity. You may have encountered these before, when you learned about rational functions with
1:46:55
holes in them, for example, the function f of x equals x minus three squared times x minus four divided by x minus four, which has a hole when x equals four, and otherwise looks like the graph of x minus three squared. This kind of discontinuity is called removable, because you could get rid of it by plugging the hole just by defining F to have an appropriate value when x equals four. So in this case, you'd want f of x to be the same when x is
1:47:30
not equal to four, but you'd want it to have the value of one when x equals four. And that would amount to plugging the hole and making it continuous. In this original example, our function had a removable discontinuity because it wasn't defined when x equals four. But a function could also have a removable discontinuity, because it's defined in the wrong place at x equals four, for example, too high or too low, to fit the trend of f. a discontinuity
1:48:02
can also occur at a vertical asymptote, where it's called an infinite discontinuity. For example, the rational function g of x is one over x minus two has an infinite discontinuity at x equals two. Occasionally, you may encounter a discontinuity is that don't look like any of these. For example, the graph of the function y equals cosine of one over X has a wild discontinuity at x equals zero, because of the wild oscillating behavior there. So for a function to be continuous
1:48:42
at x equals a, we need it to avoid all of these problems. To avoid a jump discontinuity, we can insist that the functions limit has to exist at x equals a way to avoid a hole or removable discontinuity, we can insist that F has to be defined at x equals A to avoid the other kind of removable discontinuity in which f of a is defined, but it's in the wrong place. We can insist that the limit of f of x as x goes to a has to equal f of
1:49:18
a. Sometimes the definition of continuity is written with just the third condition, and the first two conditions are implied. Notice that these three conditions not only exclude jump this continuity and removal this continuity is they also exclude infinite discontinuities and wild discontinuities. For example, in our third example, the function can't be continuous at x equals two, because it fails to have a limit at x equals two and it fails to have
1:49:46
a value at x equals two. In our wild, this continuity example, the limit also fails to exist at x equals zero, so the function can't be continuous there. So what are the places where this function f is not in us and why, please take a moment to think about it for yourself. The functions not continuous at negative three, because the function is simply not defined there, the functions not continuous at x equals one, because of that jump discontinuity.
1:50:18
In the language of limits, we say that the limit of f of x does not exist, when x equals two, the limit of the function exists and equals three, but the value of the function is down here at negative one. So the function is not continuous, because the limit doesn't equal the value at x equals three, the function is not continuous, because once again, the limit doesn't exist. Notice that x equals negative two. Even though the function turns
1:50:52
a corner, the function still continuous. Because the limit exists and equals two, and the value of the function is also two. The function drawn here is not continuous at x equals negative two, because the limit doesn't exist at x equals negative two, the limit from the left is one, while the limit from the right is zero. But it is true that the value of the function at x equals negative two is equal to the limit of the function from the left
1:51:27
side. That is, f of negative two is equal to the limit as x goes to negative two from the left of f of x. Notice that we can't say the same thing about the right limit. The limit from the right is zero, while the value of the function is one, and those are equal. In this situation, we say that f is continuous from the left, but not from the right. By the same reasoning at x equals one, the function is not continuous, but it is continuous from the right, because
1:52:04
the limit from the right is equal to the value of the function. Notice that f is not continuous from the left here, because the limit from the left is about one and a half, while the value of the function is one. In general, we say that a function f is continuous from the left at x equals j. If the limit as x goes to a from the left of f of x is equal to f of a, and a function is continuous from the right at x equals a. If the limit as x
1:52:38
goes to a from the right of f of x equals f of a, in practical terms of function is continuous from the left if the endpoint is included on the left piece, and a function is continuous from the right, if the endpoint is included on the right piece. This video gave a precise definition of continuity at a point in terms of limits. Namely, a function is continuous at the point x equals a. If the limit as x goes to a of the function is equal to the functions value at A.
1:53:18
In a previous video, we gave a definition for continuity at a point. In this video, we'll discuss continuity on an interval and continuous functions. We say that a function f of x is continuous on the open interval BC, if f of x is continuous at every point in that interval. For x to be continuous on a closed interval BC, we require it to be continuous on every point in the interior of BC, but we just require it to be continuous
1:53:54
from the right at B. And from the left at sea. We can also talk about f being continuous on half open intervals. For example, on half open interval, BC, which is open at B and closed at sea, or the other way around, or the happy to open it all from B to infinity and so on. In all of these cases, we require F to be continuous on the interior of the interval, and left or right continuous on the closed endpoints of the interval as appropriate.
1:54:27
So on what intervals is this function geovax continuous? Well, it's continuous. On this part, the arrows indicate it keeps on going. So I'd say it's continuous from negative infinity to negative one, not including the endpoint negative one. It's also continuous here, and we can include the endpoint this time. So this is from negative one to one and Then again, on this last section, we can't include one. It's not continuous there. It's not even
1:55:09
defined there. So what kinds of functions are continuous? What kinds of functions are continuous everywhere? And by everywhere, I mean, on the entire real line negative infinity infinity? Well, polynomials are a great example. Also, sine x and cosine x, the absolute value of x is another common example. There are certainly many other functions that are continuous on the whole real line. I'll let you see if you can come up with some more examples. Now,
1:55:43
if we ask the second question, what kinds of functions are continuous on their domains, we get a lot more answers, not only polynomials, but also all rational functions, things like f of x equals 5x minus two over x minus three squared times x plus four is a good example of a rational function, even though it's not continuous everywhere. Because it's not continuous when x equals three or negative four, it is continuous on its whole domain, because three
1:56:13
and four are not in the domain of this rational function. In addition, all trig functions, inverse trig functions, log and either the x functions. And pretty much all the functions we normally encounter are continuous on their domains, although their domains are not necessarily the whole real line. For example, for natural log of x, the domain is just zero, infinity, and that's where the function is continuous. In addition, the sums, differences, products
1:56:45
and quotients of continuous functions are continuous on their domains, so for example, y equals sine of x plus the natural log of x is continuous where it's defined, the compositions of continuous functions are continuous on their domains. So for example, the function y equals ln of sine of x is continuous, where defined turns out to be a bunch of disjoint intervals where sine is positive. Since continuity is defined in terms of limits, it's sometimes
1:57:24
possible to use our knowledge of which functions are continuous to calculate limits. For example, if we want to find the limit as x goes to zero of cosine of x, because cosine is continuous, we can evaluate this limit just by plugging in zero for x. And cosine of zero is one. We're using the definition of continuity here to say that the limit of the function is equal to the value of the function. The second example is a little trickier, because the function
1:57:56
inside is not continuous at x equals two. In fact, it's not defined at x equals two. But as X approaches 2x squared minus four over 2x minus four times pi, can be rewritten as x plus two times x minus two over two times x minus two times pi, which is the same thing as x plus two over two pi. For x not equal to two. So as X approaches two, this expression here, approaches two plus two over two times pi, which is just two pi. In other words,
1:58:43
the limit as x goes to two of x squared minus four over 2x minus four pi is just two pi. And therefore, the limit as x goes to two of cosine of this expression is just cosine of two pi, which is again equal to one. We're using here the fact that cosine is continuous, and a property of continuous functions, which says that the limit as x goes to a of f of g of x is equal to f of the limit as x goes to a of g of x. If f is a continuous function.
1:59:27
In other words, for continuous functions, you can pass the limit inside the function. That's all for continuity on intervals and continuous functions. The intermediate value theorem says that if f is a continuous function, on the closed interval a b and if n is any number, in between F of A and F of Bay, n f has to achieve that value and somewhere. In other words, if n is a number between F of A and F of b, then there has to be a number
2:00:11
c in the interval a, b, such that f of c equals n. In our example, there are three such possible values for C, it could be right here, since f of that number equals n, or it could be here, or here, I'll just mark the middle one. The intermediate value theorem can only be applied to continuous functions. If the function is not continuous, then it might jump over and, and never achieve that value. When application of the intermediate value theorem is to prove
2:00:56
the existence of roots or zeros of equations, we call that a real root of p of x is a real number c, such that P of C is zero, we're going to want to apply the intermediate value theorem with n equal to zero. Our polynomial is defined on the whole real line, not just an interval. But the trick here is to pick an interval a b, so that P of A is negative, and P of B is positive, or vice versa. So that the intermediate value theorem will tell us that
2:01:38
P has to pass through zero in between. I'm just going to use trial and error here and calculate a few values of p. So P of zero is easy to calculate, P of zero is just seven. P of one is going to be five minus three minus 12 plus seven, which equals negative three. So in this very lucky example, the first two numbers that we pick will work for our A and B, so we can just let A be equal 01. Because P of zero is a positive number, and P of one
2:02:19
is a negative number. So actually, the graph should look a little different. The graph looks more like this. But in any case, by the intermediate value theorem, there has to be a number c, in between, in this case, zero and one, where our polynomial p achieves this intermediate value of zero. And that number See, we don't know what it is, but we know it somewhere in the interval zero to one, that value c gives us a real root
2:02:51
for our polynomial. There may be other real roots, but we've proved there exists at least one. The intermediate value theorem has lots of other applications besides finding roots. For example, suppose you have a wall that runs in a circle around the castle, and the height of the wall varies continuously as a function of the angle. Surprisingly, the intermediate value theorem can be used to show there must be somewhere two diametrically
2:03:21
opposite places on the wall with exactly the same height. So if you can figure out a way to show this in this video, we stated the intermediate value theorem, which holds for continuous functions, and talked about a couple of applications. This video introduces the trig functions, sine, cosine, tangent, secant, cosecant, and cotangent. For right triangles. For a right triangle with sides of length A, B and C angle theta as drawn, we define
2:04:03
sine of theta as the length of the opposite side over the hypothesis. The side that's opposite to our angle theta has measure a and I have partners is this side here with measure C. So that would be a oversee for this triangle. Cosine of theta is defined as the length of the adjacent side over the length of high partners. This side here is the side adjacent to theta. Of course, the high partners is also adjacent to theta, but
2:04:39
it's special as the high partners so we don't think of it as the adjacent side. So that would be B oversea. tangent of theta is the opposite side length over the adjacent side length. So that would be a over b. The pneumonic to remember this is so A toa. That's sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. And tangent is opposite over adjacent. In fact, there's a relationship between tangent and sine and cosine. Namely,
2:05:24
tangent of theta is equal to sine of theta over cosine of theta. If you want to see why that's, that's because sine of theta over cosine of theta is given by sine, which is opposite over hypotenuse, divided by cosine, which is adjacent over hypotenuse. If we compute these fractions by flipping and multiplying, the high partners, length cancels, and we just get opposite over adjacent, which is by definition, tangent of theta. There are three more trig functions that are defined
2:06:07
in terms of sine, cosine and tangent. First of all, their sequence of theta, by definition, that's one over cosine of theta. So it's going to be one over the adjacent over hypotenuse, which is the high partners over the adjacent, which in this triangle is C over B. cosecant of theta is defined as one over sine theta. So that's one over the opposite over the hypotenuse, which is the high partners over the opposite. And for this triangle, that's going to be
2:06:44
C over a. Finally, cotangent of theta is defined as one over tan theta, that's going to be one over opposite over adjacent, flip and multiply, I get adjacent over opposite birth, which in this case is b over a. So notice that the values for cotangent cosecant and secant are the reciprocals of the values for tangent, sine and cosine, respectively. Let's use these definitions to find the exact values of all six trig functions for the angle theta
2:07:26
in this triangle. I'll start with sine of theta. That's the opposite of our hypothesis. Well, for this angle theta, the opposite side is down here, and as measured to the high partners has measured five. So sine theta is two over five. Cosine theta is adjacent over hypotenuse, but I don't know the value of this side length. But fortunately, I can find it using the Fagor in theorem says I have a right triangle here, I know that a
2:08:00
squared, I'll call this side length A plus B squared, where b is this other leg of the triangle is equal to c squared, where c is the hypothesis. So here I have a squared plus two squared equals five squared, which means that a squared plus four equals 25. A squared is 21. So A is plus or minus the square root of 21. But since I'm talking about the length of a side of a triangle, I can just use the positive answer. Returning to my computation
2:08:33
of cosine theta, I can write it as adjacent, which is the square root of 21. Over hypotony News, which is five, tangent theta is the opposite over the adjacent, so that's going to be two over the square root of 21. To compute secant of theta, that's one over cosine theta, so that's going to be one over square root of 21 over five, which is five over the square root of 21. The reciprocal of my cosine value. cosecant theta is one over sine theta, that's
2:09:17
going to be the reciprocal of my sine, so five halves, and cotangent theta is one over a tan theta, so it's going to be the reciprocal of my 10 value square root of 21 over two. Finally, we'll do an application. So if we have a kite that's flying at an angle of elevation as the angle from the horizontal bar of 75 degrees with a kite length string of 100 meters, we want to find out how high it is. I'll call the height Why? Well, we want to relate the
2:09:55
known quantities this angle and this hypothesis to The unknown quantity, the unknown quantity is the opposite side of our triangle. So if we use sine of theta equals opposite of our hypothesis, then we can relate these known amounts sine of 75 degrees to our unknown amount y, which is the opposite and are known amount of 100 meters. solving for y, this gives that y is 100 meters times sine of 75 degrees, we can use a calculator to compute
2:10:32
sine of 75 degrees, be sure you use degree mode and not radian mode, when you type in the 75. When I do the computation, I get a final answer of 96.59 meters up to two decimal places. Notice that we're ignoring the height of the person in this problem. To remember the definitions of the trig functions, you can use the pneumonic, so tau, and the fact that secant is the reciprocal of cosine cosecant the reciprocal of sine and cotangent
2:11:08
the reciprocal of tangent. In this video, I'll use geometry to compute the sine and cosine of a 30 degree angle, a 45 degree angle and a 60 degree angle. One way to compute the sine of a 45 degree angle is to use a right triangle with a 45 degree angle. This particular right triangle has I have partners of length one. Since all the angles of the triangle have to add up to 180 degrees, and we already have 90 degrees and 45 degrees,
2:11:46
the remaining angle must also be 45 degrees. So we have an isosceles triangle with two sides the same length, I'll call that side length a. If we want sine of 45 degrees, let's use this 45 degree angle here, then sine is opposite over hypotenuse. So if I can figure out how long this side length is, I'll be able to compute sine of 45 degrees. Now the Pythagorean theorem says this side length squared plus that side land squared equals
2:12:14
I have hotness squared. So we have that a squared plus a squared equals one squared. All right, that is two A squared equals one. So a squared is one half, and a is plus or minus the square root of one half. Since we're talking about the length of sides of triangles, I can just use the positive square root it's customary to rewrite this as the square root of one over the square root of two, which is one over the square root of two, and then
2:12:46
rationalize the denominator by multiplying the top and the bottom by the square root of two. That gives me a square root of two in the numerator, and the square root of two squared in the denominator, which is the square root of two over two. So the side lengths are the square root of two over two. Now I can figure out the sine of 45 degrees, by computing the opposite over the hypotenuse. That's, I'm looking at this angle. So opposite
2:13:14
is square root of two over two hypothesis one. So the sine of 45 degrees is the square root of two over two. Cosine of 45 degrees is adjacent over hypotenuse. That's this side length over this hypothesis. So that's the square root of two over two over one again, what would happen if instead of using this triangle with hypotony is one, we use this triangle, also a 4545 90 triangle with hypotony is five, not quite drawn to scale. Please
2:13:51
pause the video for a moment and repeat the computation with this triangle. This time, I'll call the side length B. Pythagoras theorem tells me b squared plus b squared equals five squared. So two, b squared equals 25. And b squared equals 25 over two, B is going to be the plus or minus the square root of 25 over two again, I can just use the positive version. And so B is the square root of 25 over the square root of two, which is five
2:14:23
over the square root of two, rationalizing the denominator, I get five root two over two. Now, sine of my 45 degree angle is opposite overhype hotness, which is five square root of two over two divided by five. That simplifies to the square root of two over two as before, and a similar computation shows that cosine of 45 degrees is also square root of two over two as before. This makes sense because sine and cosine are based on
2:15:00
ratios of sides. And since these two triangles are similar triangles, they'll have the same ratios of sides. To find the sine and cosine of 30 degrees, let's use this 30 6090 right triangle with hypotony is one. If we double the triangle, we get an angle of 30 here, so a total angle of 60 degrees here, and this angle is also 60 degrees. So we have a 60 6060 triangle, that's an equilateral triangle, all side lengths are the same. Since this
2:15:37
side length has length one, this side length is also one. This entire side length is one, which means this short side of our original triangle has length one half. Going back to my original triangle, let's use the Pythagorean Theorem to find the length of its longer side x. But agrin theorem says x squared plus one half squared equals one squared. So x squared plus a fourth equals 1x squared is three fourths. And so x is plus or minus the square root
2:16:15
of three fourths is the positive version, and get the square root of three over the square root of four, which is the square root of three over two. Now using our original triangle, again, let's compute the sine of this 30 degree angle here. We know that sine of 30 degrees is opposite of our hypothesis, the opposite of this angle is one half and the hypothesis is one. So we get a sine of one half over one, which is one half cosine
2:16:51
of 30 degrees is adjacent over hypothesis. So that's the square root of three over two divided by one. To find sine of 60 degrees and cosine of 60 degrees, we can actually use this same green triangle and just focus on this upper corner angle of 60 degrees instead. So sine of 60 degrees. Opposite overhype hotness, but this time the opposite to this angle is the square root of three over two. Cosine of 60 degrees adjacent over hypotenuse gives
2:17:32
us one half. I'll summarize the results in this table below. Notice that a 30 degree angle corresponds to pi over six radians since 30 degrees times pi over 180 is pi over six. Similarly, 45 degrees corresponds to pi over four radians and 60 degrees corresponds to pi over three radians. I recommend that you memorize the three numbers one half, root two over two, and root three over two. And the fact that one half and root three over
2:18:10
two go together. And root two over two goes with itself. From that information, it's not hard to reconstruct the triangles, you know that a 4545 90 triangle is my sauciest triangle. So it must have the side lengths were the same number goes with itself. And a 30 6090 triangle has one side length smaller than the other. So the smaller side must be one half and the larger side must be root three over two since root three is bigger than one
2:18:47
and so root three over two is bigger than one half. Doing a visual check, you can easily fill in the angles, the smaller angle must be the 30 degree angle, and the larger one must be the 60 degree one. In this video, we computed the sine and cosine of three special angles 30 degrees, 45 degrees, and 60 degrees. This video defined sine and cosine in terms of points on the unit circle. a unit circle is a circle with radius one.
2:19:26
Up to now we defined sine and cosine and tangent in terms of right triangles. For example, to find sine of 14 degrees in theory, you could draw a right triangle with an angle of 14 degrees and then calculate the sign as the length of the opposite side over the length of pi partners. But if we use this method to try to compute sine of 120 degrees, things go horribly wrong. When we draw this 120 degree angle, and this right angle, there's
2:20:04
no way to complete this picture to get a right triangle. So instead, we're going to use a unit circle, that is a circle of radius one. The figure below illustrates how right triangles and a unit circle are related. If you draw right triangles, with larger of hypotenuse one with larger and larger angles, then the top vertex sweeps out part of a unit circle. Let's look at this relationship in more detail. In this figure, I've drawn a right triangle
2:20:39
inside a unit circle, the high partners of the triangle is the radius of the circle, which is one, one vertex of the right triangle is at the origin, another vertex of the right triangle is at the edge of the circle, I'm going to call the coordinates of that vertex A, B. Now the base of this right triangle has length a, the x coordinate, and the height of the right triangle is B, the y coordinate. If I use the right triangle definition of
2:21:17
sine and cosine of theta, this right here is the angle theta, then cosine of theta is adjacent over hypotenuse, so that's a over one, or a. Notice that a also represents the x coordinate of this point on the unit circle. At angle theta from the x axis. I'll write that down. For sine of theta, if I use the right triangle definition, that's opposite overhead partners, so B over one, which is just B. But B also represents the y coordinate
2:22:00
of this point in the on the unit circle at angle theta. For tangent theta, if we use the right side triangle definition, its opposite over adjacent. So that's B over A, I can think of that as the y coordinate of the point over the x coordinate of the point. Now for angles theta, that can't be part of a right triangle, because they're too big, they're bigger than 90 degrees, like now I'll call this angle here theta, I can still use this idea of x
2:22:33
and y coordinates to calculate the sine and cosine of theta. So if I just mark this point, on the end of this line at angle theta, if I mark that to have coordinates x and y, then cosine theta, I'm still going to define as the x coordinate of this point, sine theta as the y coordinate, and tangent theta as the ratio of the y coordinate over the x coordinate. When we use this unit circle definition, we always draw theta starting from the positive
2:23:12
x axis and going counterclockwise. Let's use this unit circle definition to calculate sine, cosine and tangent of this angle fee. In our figure, we have a unit circle. And these numbers are supposed to represent the x&y coordinates of this point on the unit circle at the endpoint of this line segment, that lies at angle fee for the positive x axis. Sign fee is equal to the y coordinate cosine fee is equal to the x coordinate. And tangent of fee is given
2:23:59
by the ratio of the two, which works out to negative 0.3639 up to four decimal places. This video gives a method for calculating sine cosine and tangent in terms of the unit circle. Starting from the positive x axis, you draw the angle theta going counterclockwise. You look at the coordinates of the point on the circle where that angle ends. And the cosine of that angle theta is the x coordinate, sine of theta is the y coordinate, and tangent
2:24:43
of theta is the ratio. This video gives three properties of the trig function sine and cosine that can be deduced from the unit circle definition. Recall that the unit circle definition of sine and cosine for angle theta is that cosine theta is the x coordinate. And sine of theta is the y coordinate for the point on the unit circle at angle theta. The first property is what I call the periodic property. This says that the values of cosine and sine are
2:25:23
periodic with period two pi. And what that means is that if you take cosine of an angle plus two pi, you get the same thing as just if you took cosine of the angle. So when we write this down, we're assuming that theta is measured in radians. If we want to measure theta in degrees, the similar statement is that cosine of theta plus 360 degrees is equal to cosine of theta, we can make the same statements for sine, sine of an angle plus two pi is
2:25:58
equal to sine of the original angle, here, the angle being measured in radians. If we want to measure the angle in degrees, the statement is that sine of theta plus 360 is equal to sine of theta, we can see why this is true from the unit circle definition of sine and cosine. This is our angle theta, then theta plus two pi, the plus two pi adds a full turn around the unit circle to our angle, so we end up at the same place, theta
2:26:34
and theta plus two pi are just two different names for the same location on the unit circle. And since sine and cosine give you the y and x coordinates of that point on the unit circle, they have to have the same value. Similarly, if we consider an angle theta, and an angle theta minus two pi, the minus two pi means we go the other direction around the unit circle clockwise, we still end up in the same place. And therefore, cosine of theta minus
2:27:07
two pi, the x coordinate of that position is the same thing as cosine of theta, sine of theta minus two pi is the same thing as sine of theta, the same statements hold if we add or subtract multiples of two pi. For example, cosine of theta plus four pi is still the same thing as cosine of theta. This time, we've just gone to turns around the unit circle and still gotten back to the same place. So if we want to find cosine of five Pi, that's
2:27:39
the same thing as cosine of pi plus four pi, which is the same thing as cosine of pi. Thinking about the unit circle, pi is halfway around the unit circle. So cosine of pi means the x coordinate of this point right here. Well, that point has coordinates negative one, zero, so cosine of pi must be negative one. If I want to take sine of negative 420 degrees, well, that's sine of negative 360 degrees, minus 60 degrees, which is the same thing
2:28:17
as sine of minus 60 degrees. Thinking about the unit circle, minus 60 degrees, means I start at the positive x axis and go clockwise by 60 degrees that lands me about right here. And so that's one of the special angles that has an x coordinate of one half a y coordinate of negative root three over two. And therefore sine of negative 60 is negative root three over two the y coordinate. The next property I call the even odd property, it says that
2:28:53
cosine is an even function, which means that cosine of negative theta is the same thing as cosine of theta, while sine is an odd function, which means that sine of negative theta is the negative of sine of theta. To see why this is true, let's look at an angle theta. And the angle negative theta. A negative angle means you go in the clockwise instead of counterclockwise direction from the positive x axis.
2:29:27
The coordinates of this point by definition, r cosine theta sine theta, whereas the coordinates of this point are cosine negative theta, sine of negative theta. But by symmetry, these two points have the exact same x coordinate, and therefore cosine of theta must equal cosine of negative theta, while their y coordinates have the same magnitude, but opposite signs. This one's positive and this one's negative. Have, therefore, sine of negative theta is
2:30:04
the negative of sine of theta. Let's figure out if tan of theta isn't even or odd function. Well, we know that tan of negative theta, tangent by definition, is sine over cosine. Well, we know that sine of negative theta is the negative of sine of theta, whereas cosine of negative theta is cosine of theta. Therefore, we're getting negative sign theta over cosine theta, which is negative tan of theta. Since tan of negative theta is the
2:30:46
negative of tan of theta, tan theta is an odd function. The last property on this video is the Pythagorean property, which says that cosine of theta squared plus sine of theta squared is equal to one. A lot of times this property is written with this shorthand notation, cosine squared theta plus sine squared theta equals one. But this notation, cosine squared theta just means you take cosine of theta and square it. This property is called the
2:31:25
Pythagorean property, because it comes from the Pythagorean Theorem. Let me draw a right triangle on the unit circle. I'll call this angle theta. So the coordinates of this endpoint here are cosine theta sine theta. Since this is supposed to be a unit circle, the hypotenuse of my right triangle has length one, the base of my right triangle is just cosine theta, same thing as the x coordinate of this point. And the height of my triangle is the y coordinate
2:31:59
of the point sine theta. Now the Pythagorean theorem says that this side length squared plus that so that squared equals one squared, since one squared is the same thing as one, that gives me the Pythagorean property. But tigrayan property is handy for computing values of cosine given values of sine and vice versa. And this problem, we're told that sine of t is negative two sevenths. And T is an angle that lies in quadrant three. When we say the
2:32:33
angle lies in quadrant three, that means the terminal side of the angle lies here in quadrant three. One way to find cosine of t is to use the fact that cosine squared t plus sine squared t is equal to one. That is cosine of t squared plus negative 2/7 squared is equal to one, I can write this as cosine of t squared plus 4/49 is equal to one and so cosine of t squared is equal to one minus 4/49, which is 4540 nights. taking the square root of both sides,
2:33:21
that goes for the cosine t is plus or minus the square root of 45 over 49, that's plus or minus the square root of 45 over seven. Now since we're in the third quadrant, we know that cosine of t, which represents the x coordinate of this point, must also be negative. Therefore, cosine of t is going to be negative square to 45 over seven. It's also possible to solve this problem using the Pythagorean theorem for right triangles directly. If we
2:33:57
look at the fact that sine of t is negative two sevenths and ignore the negative sign for now, we can think of this information as telling us that we have a right triangle angle theta, whose opposite side is two, and whose high partners is seven. We call this side here a them but Tiger in theorem says us a squared plus two squared is seven squared. So a squared plus four is 49. So a squared is 45. And a is plus or minus
2:34:31
the square root of 45. Since I'm worrying about a triangle, I'm going to use the positive value. Now, cosine of t is going to be adjacent over hypotenuse. So that's going to be the square root of 45 over seven. Now I go back to thinking about positive and negative signs. And I noticed that since I'm in the third quadrant, my co sign is be negative, so I just stick a negative sign in front. This alternative solution, this is many of the
2:35:07
same ideas as the previous solution, and ultimately gets us the same answer. This video gives three properties of trig functions, the periodic property, the even odd property, and the Pythagorean property. This video is about the graphs of sine and cosine. I want to graph the functions y equals cosine t and y equals sine t, where t is in radians, I'm gonna think of this being the t axis, and this being the y axis. One way to do this is to plot points. So I'll
2:35:43
fill in this chart, using my knowledge of special angles on the unit circle. These points will be easier to graph, if I convert them all to decimals. Now plot the points for cosine and connect the dots to get a graph of y equals cosine t from t equals zero to t equals two pi. To continue the graph for t values less than zero or bigger than two pi, I could plot more points. Or I could just use the fact that the cosine values repeat. If I add or
2:36:17
subtract two pi to the my angle T, I'll be at the same place on the unit circle. So my cosine will be exactly the same. Therefore, my values of cosine, which are represented by my y values on this graph, repeat themselves. For example, when my T value is two pi plus pi over six about like here, it's cosine is the same as the cosine of just pi over six. So I'll take this dot here and repeat it over here. Similarly, the when t is like two pi
2:36:51
plus pi over four, I get the same value of cosine is when it's just pi over four. So this.is going to repeat. And I can continue repeating all my dots. This one repeats over here at two pi plus, say pi over three. And so my whole graph will repeat something like this. It also repeats on this side, something like this. Since subtracting two pi from my t values will also give me the same value of cosine. We can also plot points to get
2:37:27
a graph for sine and extend it by repetition. Going forward, I'll usually write the function sine and cosine as y equals cosine of x and y equals sine of x. When I read it this way, notice that x now refers to an angle, while y refers to a value of cosine, or sine. That's a different meaning of x and y, compared to when we're talking about the unit circle, where x refers to the cosine value, and y refers to the sine value. Now let's look at
2:38:03
some properties of the graphs of sine and cosine. The first thing you might notice is that the graph of cosine and the graph of sine are super similar to each other. In fact, you can think of the graph of cosine as just being the graph of sine shifted to the left by pi over two. So we can write cosine of x as the sine function of x plus pi over two, since adding pi over two on the inside, move the graph horizontally to the left by pi over
2:38:39
two. Or we can think of the graph of sine as being constructed from the graph of cosine by shifting the cosine graph right by pi over two, that means we can write sine of x as equal to cosine of x minus pi over two, since subtracting pi over two on the inside, shifts the cosine graph to the right by pi over two. Next, let's look at domain and range. The domain of sine and cosine is all real numbers. All right, that is negative infinity to infinity, but the range is just from negative one to
2:39:14
one. That makes sense, because sine and cosine come from the unit circle. The input values for the domain come from angles. And you can use any numbers and angle positive or negative as big as you want, just by wrapping a lot of times around the circle. The output values for the range, that is the actual values of sine and cosine come from the coordinates on the unit circle. And those coordinates can't be any bigger than one or any smaller
2:39:40
than negative one. So that gives us a range. As far as even an odd behavior, you can tell from the graph. Here's cosine, that it's symmetric with respect to the y axis and so it must be even. Whereas the graph of sine is symmetric with respect to the origin and must be odd. The absolute maximum value have these two functions is one and the absolute minimum value is negative one. We can also use the words midline amplitude and period to describe
2:40:12
these two functions. The midline is the horizontal line, halfway in between the maximum and minimum points. Here, the midline is y equals zero, the amplitude is the vertical distance between a maximum point and the midline. You can also think of the amplitude as the vertical distance between a minimum point and the midline, or as half the vertical distance between a midpoint and a max point. For the cosine function and the sine function, the amplitude is one. A
2:40:48
periodic function is a function that repeats at regular horizontal intervals. The horizontal length of the smallest repeating unit is called the period for Y equals cosine of x, the period is two pi. Notice that the period is the horizontal distance between successive peaks, or maximum points, or between successive troughs, or minimum points. algebraically, we can write cosine of x plus two pi equals cosine of x and sine of x plus two pi equals sine of x
2:41:28
to indicate that the functions repeat themselves over an interval of two pi and have a period of two pi. In this video, we graphed y equals cosine of x and y equals sine of x. and observe that they both have a midline at y equals zero, an amplitude of one and a period of two pi. sine u sort of functions are functions that are related to sine and cosine by transformations like stretching and shrinking and shifting. This video is about graphing these functions.
2:42:10
Let's start by graphing the function, y equals three sine of 2x. This function is related to the function y equals sine x. So I'll graph that first. Now, the three on the outside stretches this graph vertically by a factor of three, while the two on the inside compresses that horizontally by a factor of one half. If instead I want to graph y equals three sine 2x plus one, this plus one on the outside shifts everything up by one unit. Let's compare
2:42:48
the midline amplitude and period of our original y equals sine x are transformed y equals three sine 2x. And our further transformed y equals three sine 2x plus one, the original sine has a midline at y equals zero, an amplitude of one and a period of two pi. For the transformed function, y equals three times sine of 2x. The two on the inside shrinks everything horizontally by a factor of one half. So it changes the period of two pi into a period of one half times two pi, which is pi.
2:43:38
Since the two on the inside only affects x values and horizontal distances, it doesn't affect the midline, which is a y value, or the amplitude, which is a vertical distance. But the three on the outside does affect these things. Well, in particular, it affects the amplitude, since everything is stretched out vertically by a factor of three, the amplitude of one get stretch to an amplitude of three. In this case, the midline doesn't actually
2:44:08
change, because multiplying a y value of zero by three is still a y value of zero. Now on the third function, we've taken the second function and added one on the outside, so that shifts everything up by one. Therefore, instead of having a midline at y equals zero, we now have a midline at y equals one. The amplitude doesn't change though it's still three because shifting everything up by one doesn't affect the distance between the mid
2:44:38
mind and the end the maximum point. Also, the period is still pi since the period is a horizontal measure, and adding one on the outside only affects vertical things. Now next, let's graph the function y equals three times sine of two times quantity x minus pi over four. This function is very closely related to the First function we graphed on the previous page, that was y equals three sine of 2x. In fact, if we give the name f of x to that
2:45:11
function, and maybe we can call g of x, this other function, then we can get g of x by taking f of x and plugging in x minus pi over four in for x. In other words, g of x is f of x minus pi over four. This relationship gives me an idea for graphing g of x, the function we want to graph, we can first graph f of x, we already did that on the previous page. And then we can shift its graph to the right by pi over four, because that's what
2:45:52
you do when you subtract a number on the inside of a function. So here's the graph of y equals three sine 2x. Recall that it's just the graph of sine stretched vertically by a factor of three, and shrunk horizontally by a factor of one half. Now, to graph the function that I want, I'm going to shift this graph over by pi over four to the right. Notice that since I had my function written in factored form, I could just read off the horizontal
2:46:25
shift. But if I had written it instead, as y equals three sine 2x minus pi over two, which is algebraically equivalent, it would be easy to get confused and think that I needed to shift over by pi over two. So it's best to factor first, before figuring out what the shift is, we're factoring out the coefficient of x. If instead, we wanted to graph this function, same as the one we just graphed, it's just with a minus one on the outside,
2:47:00
that minus one would just bring everything down by one. Let's take a moment to look at midline amplitude, and period for the original parent function, y equals sine of x, and our final transformed function, y equals three sine of two times quantity x minus pi over four minus one, our original sine function has midline at y equals zero amplitude of one and period of two pi. For our transform function, the three on the outside stretches
2:47:39
vertically, so it makes the amplitude three. The minus one on the outside shifts everything down by one. So it brings the midline, y equals zero, down to Y equals negative one, the two on the inside, shrinks everything horizontally by a factor of one half. So the period becomes one half times two pi, which is pi. Finally, there's a horizontal shift going on our transformed function shifts to the right, by pi over four, this horizontal shift is sometimes called
2:48:27
the phase shift. The function we just analyzed was y equals three sine 2x minus pi over four minus one, which could also be written as y equals three sine 2x minus pi over two minus one. This is a function of the form y equals a sine B x minus c plus d, where b is positive. If we have a function of this form, or the similar function with cosine in it, then we know that the midline is going to be at y equals D. That's because the original midline
2:49:12
of sine or cosine at y equals zero gets shifted up by D, we know that the amplitude is going to be a because this A multiplied on the outside stretches everything vertically by a factor of A. to be a little more accurate, we should say the amplitude is the absolute value of A in case a is negative. If a is negative, then that amounts to a vertical reflection or a reflection over the x axis. We know that the period of the original sine or cosine is two
2:49:48
pi. And we know that this factor of B amounts to a horizontal shrink by a factor of one over B or I guess it could be a horizontal stretch by a factor of one over b If b is less than one, so because we're starting with a period of two pi, and we're multiplying by one over B, our new period is going to be two Pi over B. The trickiest thing is the horizontal shift. And to get that right, I like to factor out this B for my equation.
2:50:25
So instead of writing y equals a cosine bx minus c plus d, I'm going to write y equals A cosine B times quantity x minus c over b plus d. Similarly, if it's a sine function, I write y equals a sine B times x minus quantity c over b plus d, then I can read off the horizontal shift as C over B. And that'll be a shift to the right, if C over B is positive and a shift to the left, if C over b is negative, this might seem backwards from what you're
2:51:01
used to, but it's because we have that minus sign there. So if C over B is positive, we're actually subtracting on the inside. So that shifts right, if C of b over b is negative minus a negative is actually adding something, and that's why it shifts it to the left. So as one final example, say I wanted to graph y equals 1/3, cosine of one half x plus three minus five, that would have a midline at y equals minus five, an amplitude of 1/3, a
2:51:40
period of two pi divided by one half, which is four pi, and a horizontal shift. Better rewrite this horizontal shift of six units to the left, the horizontal shift is sometimes called the phase shift. And that's all for graphs of sinusoidal functions. This video is about graphing the trig functions, tangent, secant, cotangent and cosecant. To gain an intuition for the graph of y equals tangent of x, I think it's handy to look at the slope
2:52:25
of a line at angle theta on the unit circle. The slope of this line is the rise over the run. But the rise is given by sine of theta, and the run is given by cosine of theta. So the slope is given by sine theta over cosine theta, which is simply tan of theta. So if I want to graph y equals tan of x, I can think of x as being the angle and y as being the slope of the line at that angle. Notice if the angle is zero, the slope is zero. But
2:53:18
as the angle increases towards pi over two, the slope gets bigger and bigger heading towards infinity. As the angle goes from zero towards negative pi over two, the slope is getting negative and heading towards negative infinity at exactly pi over two and negative pi over two, we have a vertical line. And so the slope is undefined. Using this information, let's graph a rough sketch of y equals tan x. Remember, we're
2:53:58
thinking of x as the angle and y as the slope, we're going to go between an angle of negative pi over two and pi over two. So we said that the slope was zero when the angle is zero, and then it heads up towards positive infinity as we go towards the angle goes towards pi over two with an undefined value at pi over two, it goes negative heading towards negative infinity as the angle heads towards negative pi over two, also with an undefined value
2:54:33
at negative pi over two. You can also verify that for angles slightly bigger than pi over two, we have the same line as for angles that are approaching negative pi over two, and therefore this picture repeats and it turns out that tangent is periodic With period not to pi, like sine and cosine, but just pi, the period of pi makes sense because if you take a line and rotate it by 180 degrees, it's the same line with the same slope, and
2:55:17
therefore has the same value of tangents. In this graph of y equals 10x, notice that the x intercepts, all right values of x have the form negative two pi, negative pi, zero, pi, two pi, etc, you can write that as pi times k, where k is an integer, that is a positive or negative whole number or zero. This makes a lot of sense because tangent of x is sine of x over cosine of x. And so you're going to get x intercepts, that's where
2:56:01
y is zero, which is where the numerator is zero, and sine x is zero, at values of the form pi, two pi, and so on. From the graph, you can see the vertical asymptotes are at values like negative three pi over two, negative pi over two, pi over two, and three pi over two, these values can be written as pi over two times k, where k is an odd integer. Again, this makes sense from the definition of tangents since vertical asymptotes, will occur where
2:56:40
the denominator is zero, and cosine x is zero, at numbers, like negative pi over two pi over two, three pi over two, and so on, the domain of tangent is the x axis for which it's defined. So that's going to be everything except for the vertical asymptotes, we can write that as x such that x is not equal to pi over two times k, for K, an odd integer. The range or the y values go all the way from negative infinity to infinity. And the period, as we
2:57:17
mentioned previously, is pi. Since the smallest repeating unit has a horizontal width of pi, to graph y equals secant x, I'm going to remember that secant is one over cosine. So if I start with a graph of cosine, I can take the reciprocal of the y values to get the graph of secant, the reciprocal of one is one, the reciprocal of zero is undefined, so I'm not going to have a value at pi over two, negative pi over two, three pi over two, or negative three
2:57:58
pi over two. When I take the reciprocal of numbers, just less than one, I'm going to get numbers just greater than one, but I would take the reciprocal of positive numbers getting close to zero, I'm going to get really big positive numbers going up towards infinity. Similarly, on the other side, over here, I have numbers close to zero, but negative, so their reciprocals will be negative numbers heading towards negative infinity. The reciprocal
2:58:30
of negative one is negative one. And similarly here, so I'm getting kind of positive and negative buckets and upside down buckets as the graph of my sequence. Notice that secant has a period of two pi, which makes sense, since cosine has a period of two pi, it has a range that goes from negative infinity to negative one inclusive, and from one to infinity. That makes sense because the range of cosine is between one and negative
2:59:04
one, and we're taking the reciprocal of those values. The domain is everything except for the vertical asymptotes. Now the vertical asymptotes are where cosine is zero. So that is at values of the form pi over two, three pi over two, etc. That's values of the form pi over 2k, where k is an odd integer. So the domain is going to be x values such that x is not equal to pi over 2k. For K and odd integer. The x intercepts of secant Well,
2:59:47
it doesn't have any, because you can't take one over something and get the numbers zero for your y value. We've seen the graph of y equals Tana x and y equals secant x. This is the graph of y equals cotangent x. It looks similar to the graph of tangent x, it's just a decreasing function instead of an increasing one, and it has its vertical asymptotes. And it's x intercepts in different places. Finally, this green graph is the graph of y equals
3:00:27
cosecant x. It's related to the graph of sine x, since cosecant is one over sine x, and in fact, if I draw the graph of sine x in between, you can see how it kind of bounces off. Because it's the reciprocal. I encourage you to memorize the general shape of these graphs, you can always figure out the details by thinking how about how they're related to the graphs of cosine of x, and sine x. This video is an introduction to solving trig
3:01:07
equations. Let's start with the equation two cosine x plus one equals zero, I want to find all the solutions in the interval from zero to two pi, and then get a general formula for all solutions, not just those in that interval. Let me start by rewriting this equation to isolate the tricky part, which is cosine of x. So I'm going to write to cosine x equals negative one, and then divide both sides by two. Now I'm looking for the angles x between
3:01:42
zero and two pi, whose cosine is negative one half. Since negative one half is one of the special values on the unit circle, I can use my knowledge of the unit circle, to see that the angle between zero and two pi must be either two pi over three, or four pi over three, my answer needs to include both of these values. There are no other spots on the unit circle whose cosine is negative one half. But there are more angles, because we
3:02:18
can always take one of these angles and add multiples of two pi to it. So if I want to find all solutions, I can take these two principles solutions, two pi over three, and four pi over three, and simply add multiples of two pi to them. For example, two pi over three plus two pi, or two pi over three minus two pi, two pi over three plus four pi, and so on. A much more efficient way to write this is to write two pi over three plus two pi
3:02:55
times k, any integer that is any positive or negative whole number or zero. Similarly, I can write four pi over three plus two pi k, to capture all solutions, based on the principal solution for pi over three by adding and subtracting multiples of two pi. This is my final solution. Next, let's look at a tricky equation involving tangent. As usual, I'm going to start out by cleaning things up and isolating the tricky part, which
3:03:33
in this case is tangent. So let me add tangent to both sides. That'll give me three, tan x equals the square root of three. And so tan x is the square root of three over three. The square root of three over three looks suspiciously similar to value the value of square to three over two, which is a special value on my unit circle. So my suspicion is that my unit circle will again help me find this value of x without a calculator. Recall
3:04:08
that tan x is sine x over cosine x. So I'm looking for angles on the unit circle between zero and two pi with a ratio of sine over cosine will give me square root of three over three, I actually only need to look in the first quadrant and the third quadrant, because those are the quadrants where a tangent is positive. And I really only need to look at angles whose either sine or cosine has a squared of three in it. So by trial and error, I can
3:04:43
see that tan pi over six, which is sine pi over six over cosine pi over six will give me one half over root three over two That's the same thing as one half times two over three, which is one over root three. If I rationalize that, I get root three over three, so that value works. If I try tan of pi over three, instead, I get root three, which is not equal to root three over three. So pi over three doesn't work. Similarly, I can
3:05:26
work out some the values in the third quadrant, and see that seven pi over six works. But four pi over three does not. So my answer to part A includes just the two values, pi over six, and seven pi over six. Now if I want to find all solutions, not just those in the interval from zero to two pi, I noticed that I can take one of these principal solutions, and add multiples of two pi to it, because that'll give me the same angle. So I get pi
3:06:06
over six plus two pi k, and pi over six, sorry, seven pi over six plus two pi, K, any integer. This is a correct answer. But it's not as simple as it could be. Notice that seven pi over six over here on the unit circle is exactly pi more than pi over six. So instead of taking both of these and adding multiples of two pi to them, I could get all the same answers by just taking one of them and adding multiples of pi to it. So a more efficient answer is
3:06:46
to say that x equals pi over six, plus pi times k, for K any integer. This will still capture all the same solutions. Because when k is even, I'll get this family solutions. And when k is odd, I'll get this family. For example, when k is one, pi over six plus one times pi is just the original seven pi over six. If you think about the fact that tangent has a period of pi, instead of two pi, it makes a lot of sense that you should be able
3:07:22
to write the solutions in this form. In this video, we solved basic trig equations by first isolating sine, or tangent, or the same thing would work with cosine. And then using the unit circle, to find principal solutions. Principal solutions are just solutions between zero and two pi. And then adding multiples of two pi to these principal solutions to get all solutions. For tangent, we noticed that it was equivalent to just use one principal
3:08:09
solution and add multiples of pi instead of two pi. In this video, we'll introduce the idea of the derivative using graphs, secant lines and tangent lines. So I have a function here, drawn in black, the function is y equals f of x. But actually here f of x equals x squared. I also have a tangent line to my function drawn in red. This tangent line is the tangent line at the point 1.5 2.25. By a tangent line, I mean
3:08:49
a line that touches the graph of my function at this one point, and heads off in the same direction as the function. Well, normally to compute the slope, we need two points. But for the tangent line, we really only have the exact coordinates of this one point, we could approximate the slope by guessing the coordinates of some other points on the red line. But in the long run, we'll end up with a more accurate estimate if we do something
3:09:13
else. So what we're going to do instead is we're going to calculate the slope of a secant line. A secant line is a line that goes through two points on my graph. So in this case, my secant line is going through my original point, and this other point at x equals three. So that's the point three, three squared or three nine. Okay, so here's my secant line. To calculate the slope of my secant line, I use the fact that slope is rise over run. Or in other words,
3:09:43
it's the change in y over the change in x. And that's going to be written in function notation. F of three minus f of 1.5 is giving me the change in y and three minus one point gives me the change in x. Wilson's f of x is x squared, this is the same thing as three squared minus 1.5 squared over three minus 1.5. So that's just nine minus 2.25, over 1.5, which ends up as 4.5. So 4.5 is the slope of this second line. Well, the idea here is
3:10:22
that the slope of my secant line is an approximation of the slope of my tangent line. But in this example, really the secant line that I've used, it's, its slope is only a very rough approximation of the tangent line, not very accurate, all. So how could I get a better approximation of the slope of my tangent line? Well, one thing I could do is I could use as my second point, instead of using this point way out here, I could use a point closer
3:10:52
to my first point. So for example, I could use the point two, f of two. In other words, the point two, four, as my second point for my second line, so let me draw that. And let me calculate its slope, which after some arithmetic is going to give me an answer of 3.5. Well, I could continue to pick second points for my secant line, closer and closer to my first point. And I should end up with more and more accurate approximations of my tangent line,
3:11:26
let me make a little chart for this. For my next secant line, I could take my second point, something pretty close to 1.5, say 1.6. And after some arithmetic, I get a slope of 3.1. To take my second point to be 1.51, that's going to give me a slope of 3.01, and so on. To write this more generally, if I take my second point as x, then the slope of my secant line is going to be given by again the rise over the run. So that's going to be f of x
3:12:00
minus f of 1.5 divided by x minus 1.5. Change in y over the change in x, that's my slope. Now, there's no reason I necessarily have to take my second point to be on the right side of my first point, I could be using stead points on the left side here. Continue with my chart, letting my second point be one, I can do the same computations to get a slope of a secant line of 2.5. Here, I could get even closer on the left, say something like
3:12:39
1.4 and get a slope of 2.9, and so on. In general, if I have a point x f of x, and I use a secant line through that point in our original point, I calculate the slope as change in y over change in x, which is going to be f of 1.5 minus f of x divided by 1.5 minus x. Actually, I can rewrite this a little bit to make it look more like the expression up here. If I multiply the numerator and denominator by negative one, then I can rewrite this as
3:13:21
f of x minus f of 1.5 divided by x minus 1.5. So that these two expressions look exactly the same. So the only difference here on my mind is that over here, I was thinking of x as being a little bit bigger than 1.5. And here I'm thinking of x as being a little smaller than 1.5. But I get the exact same expression for the slope of the secant line either way. Now, this process of picking points closer and closer to our original point from the
3:13:52
left, and from the right, should remind you of limits. And indeed, the slope of the tangent line is the limit as x goes to 1.5 of the slope of my secant lines, which are given by this expression. This quantity is so important that's given its own name, it's called the derivative of f of x at x equals 1.5. So in other words, the derivative which is written as f prime at 1.5, is the limit as x goes to 1.5 of f of x minus f of 1.5 over x minus
3:14:32
1.5. Now based on our numerical tables, for example, here, we can see that that limit seems to be heading towards three, whether X approaches 1.5, from the right or from the left. So I'll write down the answer of three. You know, it's pretty strong. If you wanted to have a really precise argument. We'd actually need to use algebra to compute this limit. Exactly using the formula for the function itself, f of x equals x squared. And we'll
3:15:07
do examples like that in a future video. But for now, the main point is just that the slope of the tangent line is the limit of the slope of the secant lines, which is given by this formula. For now, let's look at an animation that shows how the slope of our secant lines approach the slope of our tangent line. So this black curve here is the function y equals x squared. The red line is the tangent line through the point where x equals 1.5. And
3:15:37
the blue line is a secant line that goes through the point with x coordinate 1.5. And a second point with x coordinate 2.5. The points are shown here on the right. So I'm going to use this slider here and drag my second point closer to my first point. So notice how as the x coordinate, my second point gets closer and closer to 1.5. My secant line is getting closer and closer to my tangent line. So the slope of my tangent line really is the limit
3:16:07
of the slope of my secant lines, as my x coordinate of my second point goes to 1.5. So this is true, even if I start with my second point on the left instead of the right, as I drag that second point closer and closer to the first point, the slope of my secant lines gets closer and closer to the slope of that red tangent line. We saw in our example, that the slope of our tangent line, or the derivative at 1.5, was given by the limit as x goes to
3:16:38
1.5 of f of x minus f of 1.5 divided by x minus 1.5. Well, in general, the derivative of a function y equals f of x at an x value a is given by f prime of A equals the limit as x goes to a of f of x minus F of A over x minus a, the function is said to be differentiable at A. If this limit exists. In particular, both the limit from the left and a limit from the right have to exist and be equal for the function to be differentiable at x equals a.
3:17:26
There's another equivalent version of the definition of derivative that's very common and very useful. If we're looking at the graph of a function, I'm trying to calculate the slope of the secant line between the points A, F of A and x f of x, then let's introduce the letter H to be the quantity x minus a. So h represents the run, when I'm calculating the slope of this secant line, I can write H equals x minus a, or equivalently x equals
3:18:02
a plus H. And so I can rewrite the definition of derivative in terms of H as f prime of A equals the limit as x goes to a of f of a plus h minus F of A divided by h, just by substituting in this expression for x. And h for x minus a. Well as x goes to a, x minus a is going to zero. In other words, h is going to zero. So this is equivalent to the limit as h goes to zero of f of a plus h minus f of a over h. One way to think of this is that
3:18:49
we're just relabeling this point right here, as the point A plus h, f of a plus H. And the slope of the tangent line is still the limit of the rise over the run as the run goes to zero, this is the definition of derivative that we'll use most frequently going forward when we actually calculate derivatives based on the definition in future videos. But for now, let's look at some examples to practice recognizing the derivative rather than computing
3:19:20
it. So each of these following two expressions are supposed to represent the derivative of some function at some value a. So for each example, we're supposed to find the function and figure out the value of A. Now remember, we've got two definitions of derivative going on. They're both equivalent, but they look different. One of them looks like derivative of f at a is the limit as x goes to a of f of x minus F of A over x minus a. And the
3:19:49
other version is the limit as h goes to zero of f of a plus h minus f of a over h. Now, you might notice That our first expression looks more like this first definition, because x is going to some number that's not zero. And we have both x and a number in the denominator here. Whereas our second expression looks more like the second definition, we've got the age going to zero, and we've just got the age on the denominator here. Okay, so
3:20:22
let's look at this first one here. First, let's figure out what A is here. It seems like a has got to be negative one. Since x is approaching negative one. That kind of makes sense, because now here on the denominator, x plus one could be thought of as x minus negative one. So that's our x minus a with a is negative one. Okay, great. So we've got a, now we need to find an F. And we will need the numerator here to look like f of x minus
3:20:52
f of a, well, let's try the simplest thing, we can, let's try f of x equals x plus five squared, then X plus five squared is our f of x. and f of a is our f of negative one is going to be negative one plus five squared, which is 16. So that matches up perfectly, we've got our f of x here, our f of a here, and our x minus a at the bottom. That's exactly the definition of derivative done the first one. All right, now the second one. Now, again,
3:21:28
we need to figure out what a is. And we need to find figure out what f is. This part of the expression here is supposed to be f of a plus h. So I'm going to guess, make a guess here, that f of x should be three to some power, let's just try three to the X and see how that works. Now we need this nine to be f of a. So nine has to be three to the A. And the only way that can work is if a is two. So continuing on the top, we need f of a plus h, that is f of two plus h,
3:22:11
two B three to the two plus H and that actually works perfectly. If our f of x is three to the x, it's all falling into place. So we've got our f of two plus h, r f of two, and r h. And it all works where f of x being three to the x and a being two. In this video, we introduced the idea of derivative as a slope of a tangent line. And we gave two equivalent definitions of the derivative in terms of limits. We'll continue with some interpretations
3:22:51
of derivatives in the next video. This video is algebraically intensive, and is concerned with calculating derivatives using the limit definition of derivative. Well, there are actually two versions of the limit definition of derivative. And I'll mostly use this one, the limit as h goes to zero of f of a plus h minus f of a over h. If you're interested, you can try reworking the problems using the alternative definition of derivative. The
3:23:24
limit as x goes to a of f of x minus F of A over x minus a first example, find the derivative of f of x, which is one over the square root of three minus x at x equals negative one. Well, in other words, we want to find f prime of negative one. So that's the limit as h goes to zero of f of negative one plus h minus f of negative one over h. Using our definition of f, that's the limit of one over the square root of three minus negative one plus h minus
3:24:16
one over the square root of three minus negative one, all over h. Let me clean this up a bit. So this is one over the square root of, let's see, it's three minus negative one, so that's three plus one, or four minus h minus one over the square root of four over h. And I guess I can replace the square root of four with two. Now, unfortunately, I can't just evaluate this by plugging in H equals zero, because if I try that, I get one of these zero over zero indeterminate forms. You'll run across
3:24:54
these a lot when calculating derivatives by definition, it kind of makes sense because remember the context where computing slopes of secant lines as these points get closer and closer together. So our rise and our runs are both going to zero. So it makes sense, we'll get these zero or zero indeterminate forms, we have to use our algebraic tricks that we learned before, for rewriting our expression in a way that we can calculate
3:25:18
the limit. And I see two things going on here, there's square roots lurking, and there's also fractions. So it's anybody's guess which trick I might want to apply First, the trick for square roots, which would be multiplying the top and the bottom by the conjugate. And the trick for fractions, which would be adding together my fraction for the common denominator. I guess I'll try my fraction trick first. So my common denominator for my two fractions
3:25:48
here, is just the product of these two denominators. So that's the square root of four minus h times two. Let me rewrite my fractions with this common denominator. Continuing here, I get the limit of two minus the square root of four minus h over squared, a four minus h times two, all over h. instead of dividing by H, let me multiply by one over h. And let's see here, let's see if we can evaluate by plugging in H equals zero at this stage. Unfortunately,
3:26:32
when I try to plug in, I'm still getting the zero of zero indeterminate form. But I'm not out of tricks, I haven't used the conjugate trick yet. So let's try multiplying the top and the bottom by the conjugate of the top. Once I multiply out here, I'll get four plus two square to four minus h minus two square to four minus h, minus the square root of four minus h squared, that's going to cancel out nicely. And on the bottom, I have two
3:27:12
h squared of four minus h times two plus square to four minus h, I'll leave that factored for now. That's good thing, I have unlimited space here, I'm kind of need it. Now on the numerator, I'm going to get four minus four minus h, is carrying the denominator along for the ride now, oh, I see something good. I see in the numerator, we're getting a four minus four plus H. Subtracting out those fours to zero, and then cancelling out, my H is
3:27:57
that divide by each other, I think I finally got something that I can evaluate without getting a zero over zero indeterminate form. All right, so here we've got. So as h goes to zero, I'm just going to get one over two times the square root of four times two plus a squared of four, which equals 1/16. So by now, you may have forgotten what the original problem was, I think I have it, let's go back up here, we were looking for the derivative
3:28:34
of f of x, which was one over a squared of three minus x at x equals negative one, we set up the limit definition, did a bunch of algebra first adding together fractions, then using the conjugate trick, and eventually found that that derivative equaled 1/16. The algebra doesn't get much harder than this problem here. In this next example, we're asked to find the equation of the tangent line to y equals x cubed minus 3x at x equals
3:29:04
two. So the slope of the tangent line is given by the derivative, f prime of two. So let's calculate the derivative first. f prime of two is the limit as h goes to zero of f of two plus h minus f of two, all over h. So that's the limit of two plus h cubed minus three times two plus h minus two cubed minus three times two. I'm just plugging in first two plus h for x in the definition of my function, and then I'm plugging in two for x. Now all of that needs to be
3:29:50
over h. Once again, if I try to plug in zero for H in these expressions, I'm just gonna get something that all cancels out to zero at the top, and also have zero at the bottom, one of my classic zero over zero indeterminate forms. So instead, I need to use algebra to simplify things. And hope I can calculate the limit after that. So a good trick for simplifying here is to multiply out two plus h cubed multiplies out to two cubed, plus
3:30:27
three times two squared times h, plus three times two times h squared plus h cubed. I'm getting this from the formula for multiplying out of cubic which I've memorized. But you can also get it more slowly just by writing out two plus h times itself three times and, and distributing. Now I need to subtract three times two, and three times h. And finally, I needed to subtract my two cubed and then add my three times to all this over h. Now
3:31:11
if you have my terms cancel out to zero here, two cubed minus two cubed. And let's see I've got a minus three times two and a plus three times two. And I notice all the terms that are left have H's and so I'm going to factor out an H from the top from the remaining terms here. And that gives me let's say, three times two squared, so that's 12, plus six H, plus h squared minus three over h. Now, h divided by h is one. So I'm just left with the limit
3:32:00
as h goes to zero of 12 plus six h plus h squared minus three, as h goes to zero, I can just plug in H zero, and I get 12 minus three, which is nine. So my slope of my tangent line, my derivative is nine. I'm not quite done, I still need to find the equation of the tangent line, I just know that its slope is nine. So the equation of the tangent line equation of any line is something like y equals mx plus b. and here m is nine. So I have y
3:32:33
equals 9x plus b, I just need to find the intercept B. Now, usually, to find the intercept, I need a plug in a point. what point do I have here to plug in? Well, remember, we're talking about a tangent line here. So we've got the point of tangency, the point where x equals two, and the corresponding y value is y equals two cubed minus three times two, or two. So my tangent line has to go through the point to two, which means if I plug in
3:33:10
this point for x and y, I get two equals nine times two plus b, which means that B has to equal negative 16. So the equation of my tangent line then becomes y equals 9x minus 16. I found that by first calculating the derivative to get my slope, and then using the point of tangency, plugging in the x value to get the y value, and plugging Matt in to get b to finish off the equation. So in this video, we used our tricks for evaluating limits algebraically
3:33:48
to compute some derivatives, using the definition of derivative. This is pretty labor intensive. So fortunately, pretty soon, we'll learn some shortcut methods for calculating derivatives without resorting to the definition. We've seen that the derivative of a function y equals f of x at a point, x equals A represents the slope of a tangent line through the point A f of a. But if the function f of x represents some practical quantity, like distance as
3:34:20
a function of time, or fuel efficiency as a function of speed, then the derivative will also have a practical interpretation. This video is about interpreting the derivative in different contexts. One of the most famous contexts for interpreting derivatives is problems involving motion. So let's say I'm on a bike ride, heading straight north from campus. And let's suppose that y equals f of x represents my distance from campus. So x is the time and hours and y or
3:34:55
f of x is my distance and miles, the distance away from campus. Distance north of campus. It's kind of fun to see what this graph here means in terms of my bike ride. In particular, what's going on up here, where my y values reach their maximum. And what about here, where my function is constant. Please pause the video for a moment and see if you can make up a story that fits the graph. So here at the top, my distance is no longer increasing,
3:35:25
it's actually starting to decrease. So I must have turned around and be heading back towards campus again, over here, where my f of x is constant, I probably stopped at a coffee shop to take a rest or maybe I'm fixing a flat tire. Now let's get to the questions at hand. Consider these two points, three, f of three, and four F of four, we want to interpret the slope of the secant line through those two points. Slope is change in y over change in
3:36:01
x. And y here is distance and x here is time. So change in distance over change in time. Sounds a lot like speed, or more accurately, velocity. Velocity just means speed in a certain direction, and is positive. If distance is increasing, and negative if it's decreasing. Speed is the absolute value of velocity, and there's always positive or zero. So in our case, the velocity here must be negative, because our distance is decreasing. And we
3:36:37
could estimate it very roughly as about, say four minus 12 over four minus three, so about negative eight miles per hour. But what is this negative eight miles per hour refer to. Since we're looking at the change in distance, over this entire hour long interval, the slope of my secant line gives my average velocity over this interval. It doesn't give my exact velocity, and exactly three hours or exactly four hours, only my average velocity. If I
3:37:16
want to know my exact velocity at exactly three hours, I need to look instead at the slope of the tangent line at x equals three. The velocity at an exact instant of time is sometimes called the instantaneous velocity to distinguish it from the average velocity over a time interval. Let's think for a minute about why the velocity at exactly three hours is given by the slope of the tangent line. We saw in a previous video that the slope
3:37:47
of the tangent line is the limit of the slope of the secant lines More precisely, the limit as x goes to three of f of x minus f of three over x minus three. Wow, well, each of these ratios represents an average velocity on the interval from three to x. And so the limit is the limit of average velocities on tinier and tinier intervals of time, one minute, one second 100th of a second, in the limit, as the length of the time interval goes to
3:38:20
zero, we're going to get the exact velocity at exactly three hours. So to repeat, in this example, the slope of the secant line represents the average velocity over time interval, and the derivative at x equals three, written f prime of three, which is also the slope of the tangent line, that derivative represents the instantaneous velocity at x equals three. More generally, if f of x represents any quantity that's changing, then the slope of the secant
3:38:57
line represents an average rate of change. While the slope of the tangent line, f prime of a represents an instantaneous rate of change. Let's see how that works in a couple of other examples. Let's suppose that f of x represents the temperature of a cup of coffee and degrees Fahrenheit as a function of time and minutes since he said it on the counter. So let's interpret the first equation. f of zero is 140. Well, that just means that at time zero,
3:39:32
the temperature is 140 degrees. What about the equation f of 10 minus f of zero is negative 20. That's saying that the temperature goes down by 20 degrees as x the time goes from zero to 10 minutes. Now, what about this quotient here being equal to negative two? Well, this quotient looks a lot like the slope of a secant Why, right, so it must be an average rate of change. And in this context, we have the temperature is decreasing by an average of
3:40:11
two degrees per minute, as x changes from zero to 10 minutes. Finally, the derivative of f at 15 is negative point five means that at exactly 15 minutes, the temperature is decreasing at a rate of point five degrees per minute. negative numbers here always mean decreasing, and f prime is an instantaneous rate of change. Let's look at another example. Please pause the video and try this one for yourself. Here g of x represents the fuel
3:40:51
efficiency of a Toyota Prius and mpg as a function of x, the speed in miles per hour that is traveling g of 45 is 52 means that at 45 miles per hour, the fuel efficiency is 52 miles per gallon. The second statement is saying that as speed increases from 35 to 45 miles per hour, fuel efficiency goes up by 10. That's 10 miles per gallon. The third statement says that the average rate of change of fuel efficiency is two miles per gallon per mile per hour, as speed increases from
3:41:38
35 to 40 miles per hour. So going up from 35 to 40, gives you better fuel efficiency here. On the other hand, when you're going 60 miles per hour, your fuel efficiency is decreasing at a rate of two miles per gallon per mile per hour. So I bet the optimal fuel efficiency here occurs somewhere between 40 and 60 miles per hour. In this video, we've interpreted the slope of the secant line as the average rate of change, and the slope
3:42:19
of the tangent line with the derivative as an instantaneous rate of change. I hope that these general principles will help you interpret the derivative in a variety of contexts that you might encounter throughout your life. In this video, we'll think of the derivative of a function as being a function in itself, we relate the graph of a function to the graph of its derivative, and we'll talk about where the derivative does not exist. We've seen
3:42:49
that for a function f of x and a number A, the derivative of f of x at x equals A is given by this formula. But what if we let a very, if we compute f prime of a lots of different values of a, we can think of the derivative of f prime as itself being a function, I'm going to rewrite this definition of derivative with x in the place of a just to make it look a little more like standard function notation. So f prime as a function of x is the limit
3:43:20
as h goes to zero of f of x plus h minus f of x over h. This isn't anything substantially different from what we've been doing before, it's just a difference in perspective. So let's do one more example of computing the derivative by hand using the definition, but a general number x instead of a specific value, the function we're going to use is f of x equals one over x. And first, let's just write down the definition of derivative in general.
3:43:49
So f prime of x is given by this formula. And using the fact that f of x is one over x, I can rewrite this as one over x plus h minus one over x all over h. So as usual, this is a zero over zero indeterminate form. If I plug in zero for H, I'm just going to get one over x minus one over x on the numerator, which is zero. and plugging in zero for H gives me zero on the denominator two. So I'll need to use some algebra to rewrite things
3:44:23
to get an A form that I can evaluate it. Let's add together our fractions in the numerator here. The common denominator I need to use is x plus h times x. So I multiply this fraction by x over x and the next fraction by x plus h over x plus h. All that's over h. And now continuing, I get x minus x plus h over x plus h times X and instead of dividing the whole thing by h here, multiply by one over h to get the limit have x minus x minus h
3:45:04
over x plus h times x times h. Now I can subtract off my x's here. And after I do that, I can divide my minus h by my H, to get just a minus one on the numerator here. So that's just the limit of negative one over x plus h times x. And now I'm in a good position because I can plug in H equals zero and get something that makes sense. Namely, I'm getting a limit of negative one over x plus zero times x or negative one over x squared as my derivative.
3:45:45
In this example, we're given the graph of a function that's supposed to represent the height of an alien spaceship above the Earth's surface, we want to graph the rate of change. The rate of change means the derivative of our function, but we're not given an equation to work with. So we'll just have to estimate the derivative based on the shape of the graph by thinking about slopes of tangent lines. I'll start by drawing a new set of axes where
3:46:15
I can graph my derivative. And I'll consider my original function, which I'll call g of x, piece by piece. For x values between zero and two, my original function g of x looks like a line, it has slope negative one, since the rise is negative two, while the run is two, for any point on the straight line segment, the tangent line will also be a straight line with slope negative one, and therefore, the derivative will be negative one. For x values between zero and
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two, I'm going to ignore the time being what happens when x is exactly zero or exactly two, and just look at the interval of X values between two and three. Here, g of x is completely flat. So tangent lines at any of these points will have slope zero. And I'll draw a derivative of zero. When x is between two and three, I'll postpone worrying about the derivative when x is exactly three. And just think about the derivative when x is between three and
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five, where g of x is flat again, so it's tangent lines will have slopes of zero. And I'll draw again, a derivative is zero when x is between three and five. Now things get a little more interesting. As x increases from five to about seven, g of x is an increasing function. The slope of tangent lines here are positive, starting at about, say, a slope of three, and decreasing to a slope of zero, when x is seven, I can draw that down here.
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As x increases from seven, the tangent lines now have negative slopes, going to a maximum negative slope of about negative one here, and then heading towards a slope of zero, when x is just shy of 10. My estimates of three and negative one for the slopes of my tangent lines are just rough estimates based on approximating the rise over the run. As x increases from 10, the tangent line slope is positive, and getting steeper and steeper,
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so my derivative is going to be positive and increasing. That's the basic shape of the derivative. Now let's see what happens at these special points like 235 and zero. To figure out the derivative at x equals two, let's go back to the definition of derivative as the limit of the slopes of the secant lines. If I draw a secant line, using a point on the left, I'll just get this line that lines up with this line and has a slope of negative
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one. But if I compute the slope of a secant line, using a point on the right, I'll get a slope of zero. So the limit from the left and the limit from the right of the slopes of my secret lines will be different. And so my limit does not exist and my derivative does not exist. And so I'll just draw this as an open circle at x equals two. Next, let's look at the derivative when x equals three. Remember that the derivative at three is the
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limit as h goes to zero, of g of three plus h minus g of three over eight. Well, if H is bigger than zero, then g of three plus H is going to be about a half, because three plus H is to the right of three. On the other hand, if H is less than zero, g of three, plus H is two, because three plus H is actually a number less than three, g of three itself is equal to one half, based on the filled in bubble here. so if we calculate the limit,
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as h goes to zero from the positive side, we get the limit of one half minus one half over age, which is just the limit of zeros, so that's zero. On the other hand, if we compute the limit from the left, we get the limit of two minus a half over eight, which is the limit of three halves over h. And as h goes to zero, that limit is negative infinity. So once again, the left limit and write limit are not equal. And so the limit of the slopes
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of the secant lines does not exist, and there's no derivative at x equals three. And I'll draw an open circle there to add x equals five, again, we have a corner. And by the same sort of argument, we can conclude that the derivative does not exist. And finally, when x equals zero, we can only have a limit from the right not the left. And so by that sort of technical reason, we don't have a derivative at that left endpoint either. So we've drawn a rough graph of the rate of change
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of the height of our alien spaceship, as it comes closer to Earth beams down to pick up Earthlings and then makes us escape up to the mothership. It's interesting to observe that the domain of the original function g of x is from zero to infinity, but the domain of g prime is somewhat smaller, and just goes from zero to two, then from two to three, then from three to five, and finally, from five to infinity, missing some places where
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the function originally existed. We saw in the previous example, that the derivative doesn't necessarily exist at all the x values where the original function exists. Please pause the video for a moment and try to come up with as many different ways as you can, that a derivative can fail to exist at an x value x equals a. Well, one kind of boring way that a function can fail to have a derivative at x equals A is if f of x itself fails to exist. at x equals a, for
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example, if it has a hole, like in this picture, we saw in the previous example, with the alien spacecraft, that a derivative can fail to exist when the function turns a corner. When we tried to evaluate the derivative in that example, by taking the limit of the slope of the secant lines, the limit from the left, and a limit from the right did not agree. A famous example of a function that turns a corner is the absolute value function. For
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the absolute value function, f prime of x is negative one, since the slope here is negative one, when x is less than zero, and it's positive one when x is greater than zero, but f prime of zero itself does not exist. A function with a casp also fails to have a derivative at the cusp. In the alien spaceship example, we also saw that F can fail to have a derivative at a discontinuity. But there's another way that a derivative can fail to exist, even
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when f has no cost per corner discontinuity. Let's look at the function f of x equals x to the 1/3 graphed here, what's going on at x equals zero. At that instant, the tangent line is a vertical with a slope that's infinite or undefined. So the limit of the slopes of the secant lines will fail to exist because it'll be infinite. A function is called differentiable at x equals a, if the derivative exists at a function is differentiable on an open interval,
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if f is differentiable at every point in that interval. So all of the examples on the previous slides are examples of places where a function is not differentiable. All of these examples are important. But I'm going to focus on the example involving discontinuity. In general, if f of x is not continuous at x equals a, then f is not differentiable at x equals a. This is what we saw in the example involving the jump discontinuity. an equivalent
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way of saying the same thing is that f is differentiable at x equals a, then f has to be continuous at x previous slides are examples of places where a function is not differentiable. All of these examples are important. But I'm going to focus on the example involving discontinuity. In general, if f of x is not continuous at x equals a, then f is not differentiable at x equals a. This is what we saw in the example involving the jump discontinuity. an equivalent
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way of saying the same thing is that f is differentiable at x equals a, then f has to be continuous at x equals a. equals a. However, if all we know is that f is continuous at x equals a, then we can't conclude anything about whether or not is differentiable there, f may or may not be differentiable at x equals a. Remember the square root example, the square root of x is continuous at x equals zero, but it's not differentiable there because of the corner. In this video, we related the
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graph of a function to the graph of its derivative. By thinking about the slopes of tangent lines, we also looked at several ways that a derivative can fail to exist at a point and noted that if a function is differentiable, it has to be continuous. This video gives a proof that differentiable functions are continuous. What we want to show here is that if a function is differentiable at a number x equals a, then it's continuous at x equals a. Let me
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call the function f of x. And I'm going to start out by writing down what it means for f of x to be differentiable at x equals a. That means that the limit as x goes to a of f of x minus F of A over x minus a exists and equals this finite number that we call f prime However, if all we know is that f is continuous at x equals a, then we can't conclude anything about whether or not is differentiable there, f may or may not be differentiable at x equals
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a. Remember the square root example, the square root of x is continuous at x equals zero, but it's not differentiable there because of the corner. In this video, we related the graph of a function to the graph of its derivative. By thinking about the slopes of tangent lines, we also looked at several ways that a derivative can fail to exist at a point and noted that if a function is differentiable, it has to be continuous. This video gives a proof that
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differentiable functions are continuous. What we want to show here is that if a function is differentiable at a number x equals a, then it's continuous at x equals a. Let me call the function f of x. And I'm going to start out by writing down what it means for f of x to be differentiable at x equals a. That means that the limit as x goes to a of f of x minus F of A over x minus a exists and equals this finite number that we call f prime of a. Now I'm going to multiply both sides of this equation
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of a. Now I'm going to multiply both sides of this equation by the limit as x goes to a of x minus a. Now wait a second, before I go any further on to make sure this is legit. Does this limit actually exist? Well, yeah, because the limit as x goes to a of x exists, that's just a, and the limit as x goes to a of a exists, that's a also. So the limit of the difference has to exist. And actually, that limit of
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the difference is the difference of the limits, which is just going to be a minus a or a zero. So I've actually just multiplied both sides by zero and a fancy form. by the limit as x goes to a of x minus a. Now wait a second, before I go any further on to make sure this is legit. Does this limit actually exist? Well, yeah, because the limit as x goes to a of x exists, that's just a, and the limit as x goes to a of a exists,
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that's a also. So the limit of the difference has to exist. And actually, that limit of the difference is the difference of the limits, which is just going to be a minus a or a zero. So I've actually just multiplied both sides by zero and a fancy form. This is actually a surprisingly useful thing to do. Because I have a product of a limit of two limits here, both of which exist. So by the product rule for limits, I can rewrite this as the limit as x goes to a
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This is actually a surprisingly useful thing to do. Because I have a product of a limit of two limits here, both of which exist. So by the product rule for limits, I can rewrite this as the limit as x goes to a of x minus a times f of x minus F of A over x minus a. And canceling these two copies of x minus a, which is fine to do when x is near a, just not when x equals a, I get that the limit of f of x minus f of a is equal to this limit over here. Well, we said this
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limit was just zero. So my limit on the left is equal to zero. of x minus a times f of x minus F of A over x minus a. And canceling these two copies of x minus a, which is fine to do when x is near a, just not when x equals a, I get that the limit of f of x minus f of a is equal to this limit over here. Well, we said this limit was just zero. So my limit on the left is equal to zero. And now I'm so close, I'd like to apply five the limit rule to this difference to break
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it up into a difference of limits, but I can't quite do that, because I'm not sure yet that the limit, as x goes to a of f of x exists, that's sort of part of what I'm trying to prove as far as continuity. So instead, I think I'm gonna add to both sides, a limit that I do know exists. And that's the limit And now I'm so close, I'd like to apply five the limit rule to this difference to break it up into a difference of limits, but I can't quite do that, because I'm not sure yet that
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the limit, as x goes to a of f of x exists, that's sort of part of what I'm trying to prove as far as continuity. So instead, I think I'm gonna add to both sides, a limit that I do know exists. And that's the limit of f of a. of f of a. Now, I do know that both of these two limits on the left side exists, so I can use the limit rule about sums to rewrite this limit. Now, I do know that both of these two limits on the left side exists, so I can use the limit rule about sums to rewrite this limit.
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Now, on the left side, I can cancel out my copies of f of a, whatever number that is, and I get that the limit as x goes to a of f of x which has to exist by the the limit rule for psalms that I applied Now, on the left side, I can cancel out my copies of f of a, whatever number that is, and I get that the limit as x goes to a of f of x which has to exist by the the limit rule for psalms that I applied above. above.
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That limit their past to equal the limit as x goes to a of f of a, well, f of a is just some number, doesn't matter what X is doing f of a is it's just f of a, whatever that is. So this limit on the right is just f of a, and look at that. That's exactly what it means for a function to be continuous at x equals a, the limit as x goes to a of f of x equals f of a. So f is continuous at x equals a, hella proof is complete. In this video,
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we prove that if f is differentiable at x equals a, then f is continuous at x equals a. This statement is equivalent to another statement known in logic as its contrapositive, which says that if f is not continuous, at x equals a, then f is not differentiable at x equals a. In this video, we'll learn a few rules for calculating derivatives, namely, the power role, and the derivatives of sums, differences and constant multiples. These
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rules will give us shortcuts for finding derivatives quickly, without needing to resort to the old limit definition of derivative. In this video, we'll only do statements of the rules and some examples, there won't be any proofs or justification for why the rules hold. Instead, these proofs will be in a separate later video. Let's start with some basics. First of all, if we have a constant C, and we will have the function f of x equals C. So if I graph
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that, it's just going to be a straight horizontal line. That limit their past to equal the limit as x goes to a of f of a, well, f of a is just some number, doesn't matter what X is doing f of a is it's just f of a, whatever that is. So this limit on the right is just f of a, and look at that. That's exactly what it means for a function to be continuous at x equals a, the limit as x goes to a of f of x equals f of a. So f is continuous at x equals a, hella proof is complete. In this video,
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we prove that if f is differentiable at x equals a, then f is continuous at x equals a. This statement is equivalent to another statement known in logic as its contrapositive, which says that if f is not continuous, at x equals a, then f is not differentiable at x equals a. In this video, we'll learn a few rules for calculating derivatives, namely, the power role, and the derivatives of sums, differences and constant multiples. These
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rules will give us shortcuts for finding derivatives quickly, without needing to resort to the old limit definition of derivative. In this video, we'll only do statements of the rules and some examples, there won't be any proofs or justification for why the rules hold. Instead, these proofs will be in a separate later video. Let's start with some basics. First of all, if we have a constant C, and we will have the function f of x equals C. So if I graph
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that, it's just going to be a straight horizontal line. So the derivative So the derivative df dx, ought to be zero, because the slope of the tangent line of this straight line is just df dx, ought to be zero, because the slope of the tangent line of this straight line is just zero. zero. Another simple example, is the derivative of the function f of x equals x. So again, if I draw the graph, that's just going to be a straight line with slope one. And so
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the tangent line for the straight line will also have slope one. And the derivative, f prime of x has to be always equal to one. These two simple examples are actually special cases of the power rule, which is one of the most useful rules for calculating Another simple example, is the derivative of the function f of x equals x. So again, if I draw the graph, that's just going to be a straight line with slope one. And so
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the tangent line for the straight line will also have slope one. And the derivative, f prime of x has to be always equal to one. These two simple examples are actually special cases of the power rule, which is one of the most useful rules for calculating derivatives. derivatives. So the power rule says that if you have the function, y equals x to the n, where n is any real number, then you can find the derivative d y dx, simply by pulling that exponent and
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down and multiplying it in the front and then reducing the exponent by one. So for example, if you want to calculate the derivative of y equals x to the 15th, D y dX here is just going to be 15 times x to the 15 minus one, or 14. The second example, f of x equals the cube root of x might not immediately look like an example where we can apply the power rule. But if we rewrite it, by putting the cube root in exponential tation as x to the
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1/3, now we can apply the power rule, we bring the 1/3 down, multiplied on the front and So the power rule says that if you have the function, y equals x to the n, where n is any real number, then you can find the derivative d y dx, simply by pulling that exponent and down and multiplying it in the front and then reducing the exponent by one. So for example, if you want to calculate the derivative of y equals x to the 15th, D y dX here is just
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going to be 15 times x to the 15 minus one, or 14. The second example, f of x equals the cube root of x might not immediately look like an example where we can apply the power rule. But if we rewrite it, by putting the cube root in exponential tation as x to the 1/3, now we can apply the power rule, we bring the 1/3 down, multiplied on the front and reduce the reduce the exponent of 1/3 by one, or 1/3 minus one is negative two thirds. So we found the derivative
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here using the power rule, we could rewrite it if we want to using exponent rules, as one over 3x to the two thirds, either answers good. exponent of 1/3 by one, or 1/3 minus one is negative two thirds. So we found the derivative here using the power rule, we could rewrite it if we want to using exponent rules, as one over 3x to the two thirds, either answers good. In the third example, g of x is one over x to the 3.7. Again, we need to do a little
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rewriting before we can apply the power rule. So I'm going to rewrite g of x as x to the minus 3.7. Using exponent rules, now I can find dg dx by pulling down the negative 3.7 multiplying in the front, and now I have to reduce negative 3.7 by one, so I subtract one that gives me x to the negative 4.7. Again, I can rewrite this if I wish as negative 3.7 over x to the 4.7. It's important to notice that in all these examples and in fact, in
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any example where the power rule applies, the variable x is in the base. And the exponent is just a constant, just a real number. The constant multiple rules says that if c is just a constant real number, and f is a differentiable function, then the derivative of C times f of x is just c times the derivative of f of x. In other words, when we take the derivative, we can just pull a constant outside of the derivative sign. Let's use this rule in an
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example. If we want to take the derivative of 5x cubed, that's the same thing as five times the derivative of x cubed. And now using the power rule, we can bring down the three and get 15x squared. f and g are differentiable functions, then the derivative of f of x plus g of x is the derivative of f plus the derivative of g. Similarly, for a difference, if f and g are differentiable functions, then the derivative of the difference is just the difference of
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the derivatives. Now let's use all these rules together to calculate the derivative of this polynomial. To find the y dx, we can use the sum and difference rule to calculate the derivative of each term separately. Now using the constant multiple rule and the power rule, we can bring out the seven, bring down the three getting x squared here, similarly, for the next term, five times two times x to the one, four times the derivative of x, which is just one, and
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the derivative of a constant two is just zero. So simplifying, we get 21x squared minus 10x plus four, and notice that the derivative of the original polynomial is just another polynomial of one less degree. In this video, we use some shortcuts to calculate the derivatives of various functions, especially polynomials. If you're interested in seeing where these rules come from, how they're derived from the limit definition of derivative, then look
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for another video coming soon on proofs. This video is about identities involving trig functions like sine and cosine. But I want to start with some examples that just involve In the third example, g of x is one over x to the 3.7. Again, we need to do a little rewriting before we can apply the power rule. So I'm going to rewrite g of x as x to the minus 3.7. Using exponent rules, now I can find dg dx by pulling down the negative 3.7
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multiplying in the front, and now I have to reduce negative 3.7 by one, so I subtract one that gives me x to the negative 4.7. Again, I can rewrite this if I wish as negative 3.7 over x to the 4.7. It's important to notice that in all these examples and in fact, in any example where the power rule applies, the variable x is in the base. And the exponent is just a constant, just a real number. The constant multiple rules says that if c is
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just a constant real number, and f is a differentiable function, then the derivative of C times f of x is just c times the derivative of f of x. In other words, when we take the derivative, we can just pull a constant outside of the derivative sign. Let's use this rule in an example. If we want to take the derivative of 5x cubed, that's the same thing as five times the derivative of x cubed. And now using the power rule, we can bring down the three
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and get 15x squared. f and g are differentiable functions, then the derivative of f of x plus g of x is the derivative of f plus the derivative of g. Similarly, for a difference, if f and g are differentiable functions, then the derivative of the difference is just the difference of the derivatives. Now let's use all these rules together to calculate the derivative of this polynomial. To find the y dx, we can use the sum and difference rule to calculate the derivative
4:18:17
of each term separately. Now using the constant multiple rule and the power rule, we can bring out the seven, bring down the three getting x squared here, similarly, for the next term, five times two times x to the one, four times the derivative of x, which is just one, and the derivative of a constant two is just zero. So simplifying, we get 21x squared minus 10x plus four, and notice that the derivative of the original polynomial is just another
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polynomial of one less degree. In this video, we use some shortcuts to calculate the derivatives of various functions, especially polynomials. If you're interested in seeing where these rules come from, how they're derived from the limit definition of derivative, then look for another video coming soon on proofs. This video is about identities involving trig functions like sine and cosine. But I want to start with some examples that just involve
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quadratic functions. If I want to find the solutions to this equation, I can rewrite it x squared minus 6x minus seven equals zero, factor it, x minus seven times x plus one equals zero, set the factors equal to 0x minus seven equals zero, or x plus one equals zero. And that gives me the solutions, x equals seven, or x equals negative one. Next, let's look at this more complicated equation. I'm going to try to solve that for x by multiplying
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out the right hand side. Next, our combined terms on the right hand side. So that gives me x squared minus 6x on both sides, well, x squared minus 6x is equal to x squared minus 6x. That's true no matter what I plug in for x, and therefore, all values of x satisfy this equation, we can say that the solution set is all real numbers. The second equation is called an identity, because it holds for all values of the variable. The first equation,
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on the other hand is not an identity, because it only holds for some values of x and not all values. Please pause the video for a moment and try to decide which of the following three equations or identities that is, which of these equations hold for all values of the variable. To start out, you might want to test them by plugging in a few values of the variable and see if the equation holds. The first equation is not an identity. It does
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hold for some values of x. For example, if x equals zero, then sine of two times zero is zero. Two times sine of zero is also zero. So it does hold when x is zero. However, when x is say pi over two, then sine of two times pi over two, that's the same thing as sine of pi, which is zero, but two times sine of pi over two, that's two times one, or two, and zero is not equal to two. So the equation does not hold for x equals pi
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quadratic functions. If I want to find the solutions to this equation, I can rewrite it x squared minus 6x minus seven equals zero, factor it, x minus seven times x plus one equals zero, set the factors equal to 0x minus seven equals zero, or x plus one equals zero. And that gives me the solutions, x equals seven, or x equals negative one. Next, let's look at this more complicated equation. I'm going to try to solve that for x by multiplying
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out the right hand side. Next, our combined terms on the right hand side. So that gives me x squared minus 6x on both sides, well, x squared minus 6x is equal to x squared minus 6x. That's true no matter what I plug in for x, and therefore, all values of x satisfy this equation, we can say that the solution set is all real numbers. The second equation is called an identity, because it holds for all values of the variable. The first equation,
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on the other hand is not an identity, because it only holds for some values of x and not all values. Please pause the video for a moment and try to decide which of the following three equations or identities that is, which of these equations hold for all values of the variable. To start out, you might want to test them by plugging in a few values of the variable and see if the equation holds. The first equation is not an identity. It does
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hold for some values of x. For example, if x equals zero, then sine of two times zero is zero. Two times sine of zero is also zero. So it does hold when x is zero. However, when x is say pi over two, then sine of two times pi over two, that's the same thing as sine of pi, which is zero, but two times sine of pi over two, that's two times one, or two, and zero is not equal to two. So the equation does not hold for x equals pi over two. over two.
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Since it doesn't hold for all values of the variable, it's not an identity. The second equation is an identity. You can build some evidence for this by plugging in numbers. For example, cosine of zero plus pi, which is negative one is the same thing as negative of cosine of zero. You can also check for example, that cosine of pi over six plus pi is the same thing as negative cosine of pi over six. But even if we check a zillion examples,
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that's just evidence, it's not a proof that the identity holds, we could have just gotten lucky with the values we picked, we can build a little bit stronger evidence, by looking at graphs, I'm going to put theta on the x axis in a pie graph, y equals cosine of theta plus Since it doesn't hold for all values of the variable, it's not an identity. The second equation is an identity. You can build some evidence for this by plugging in numbers.
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For example, cosine of zero plus pi, which is negative one is the same thing as negative of cosine of zero. You can also check for example, that cosine of pi over six plus pi is the same thing as negative cosine of pi over six. But even if we check a zillion examples, that's just evidence, it's not a proof that the identity holds, we could have just gotten lucky with the values we picked, we can build a little bit stronger evidence, by looking
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at graphs, I'm going to put theta on the x axis in a pie graph, y equals cosine of theta plus pi, pi, that's just like the graph of cosine shifted over to the left by pi. On the other hand, if I graph y equals negative cosine theta, that's the graph of cosine theta, reflected across the x axis, which gives us the exact same graph. So graphing both sides gives us strong evidence that this equation is an identity, it holds for all values of theta. Now, the
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strongest evidence of all would be an algebraic proof, which we'll do later in the course, once we have a formula for the cosine of a sum of two angles. that's just like the graph of cosine shifted over to the left by pi. On the other hand, if I graph y equals negative cosine theta, that's the graph of cosine theta, reflected across the x axis, which gives us the exact same graph. So graphing both sides gives us
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strong evidence that this equation is an identity, it holds for all values of theta. Now, the strongest evidence of all would be an algebraic proof, which we'll do later in the course, once we have a formula for the cosine of a sum of two angles. In the meantime, let's look at equation C. It turns out equation C is an identity. And we could build evidence for it again by plugging in values for x, or by graphing the left side
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and the right side separately, and checking to see that the graphs coincided. But for this example, I'm going to go ahead and do an algebraic verification. In the meantime, let's look at equation C. It turns out equation C is an identity. And we could build evidence for it again by plugging in values for x, or by graphing the left side and the right side separately, and checking to see that the graphs coincided. But for this example, I'm going to go ahead and do an algebraic verification.
4:29:02
In particular, I'm going to start with the left side of the equation, and rewrite things and rewrite things until I get to the right side of the equation. The first thing I'll rewrite In particular, I'm going to start with the left side of the equation, and rewrite things and rewrite things until I get to the right side of the equation. The first thing I'll rewrite is secant and tangent in terms of their constituent functions, sine and cosine. Since secant of
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x is one over cosine x, and tangent of x is sine x over cosine x, I can rewrite this expression as one over cosine x minus sine x times sine x over cosine x. I can clean up those fractions and write this as one over cosine x minus sine squared x over cosine x. Now, I noticed that I have two fractions with the same denominator. So I can pull them together as one minus sine squared x over cosine x. Next, I'm going to rewrite the numerator one minus sine squared
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x using the Pythagorean identity that says that cosine squared x plus sine squared x equals one, and therefore, one minus sine squared x is equal to cosine squared x just by subtracting sine squared x from both sides. So I can replace my numerator, one minus sine squared x with cosine squared x. And canceling one cosine from the top and from the bottom, that's the same thing as cosine of x, which is the right hand side that I was trying to
4:31:29
get to. So a combination of a bunch of algebra, and the Pythagorean identity allows me to prove that this equation is true for all values of x, is secant and tangent in terms of their constituent functions, sine and cosine. Since secant of x is one over cosine x, and tangent of x is sine x over cosine x, I can rewrite this expression as one over cosine x minus sine x times sine x over cosine x. I can clean up those fractions
4:32:20
and write this as one over cosine x minus sine squared x over cosine x. Now, I noticed that I have two fractions with the same denominator. So I can pull them together as one minus sine squared x over cosine x. Next, I'm going to rewrite the numerator one minus sine squared x using the Pythagorean identity that says that cosine squared x plus sine squared x equals one, and therefore, one minus sine squared x is equal to cosine squared x just
4:33:01
by subtracting sine squared x from both sides. So I can replace my numerator, one minus sine squared x with cosine squared x. And canceling one cosine from the top and from the bottom, that's the same thing as cosine of x, which is the right hand side that I was trying to get to. So a combination of a bunch of algebra, and the Pythagorean identity allows me to prove that this equation is true for all values of x, it's an identity. The best way to prove that an equation is an identity
4:33:42
it's an identity. The best way to prove that an equation is an identity is to use algebra and to use other identities, like the Pythagorean identity to rewrite one side of the equation till it looks like the other side. The best way to prove the net equation is not an identity is to plug in numbers that break the identity. That is make the equation not true. Now if you're just trying to decide if an equation has an identity
4:34:18
or not, and not worried about proving it, then I recommend plugging in numbers, or graphing the left and right sides to see if those graphs are the same. Recall that an identity is an equation that holds for all values of the variable. This video states and proves three identities called the Pythagorean identities. The first one is the familiar cosine squared theta plus sine squared theta equals one. The second one says tan squared theta plus
4:34:55
one equals secant squared theta. And the third one goes cotangent squared theta plus one equals cosecant squared theta. Let's start by proving that cosine squared theta plus sine squared theta equals one. I'll do this by drawing the unit circle with a right triangle inside it by the definition of sine and cosine, the x&y coordinates of this top point, r cosine theta and sine theta, the high partners, my triangle is one, since that's the radius of
4:35:25
my unit circle. Now the length of the base of my triangle is the same thing as the x coordinate of this point. So that's equal to cosine theta. The height of this triangle is the same thing as the y coordinate of this point. So that's sine theta. Now the Pythagoras theorem for right triangles, says this side length squared plus that side length squared is equal to the hypothenar squared. So by the Pythagorean theorem, we have that cosine
4:35:59
theta squared plus sine theta squared equals one squared, I can rewrite that as cosine squared theta plus sine squared theta equals one, since one squared is one, and cosine squared theta is just a shorthand notation for cosine theta squared. That completes the proof of the first Pythagorean identity, at least in the case, when the angle theta is in the first quadrant. In the case, when the angle was in a different quadrant, you can
4:36:33
use symmetry to argue the same identity holds. But I won't give the details here. To prove the next Pythagorean identity, tan squared theta plus one equals secant squared theta, let's use the first without your an identity, which said that cosine squared theta plus sine squared theta equals one, I'm going to divide both sides of this equation by cosine squared theta. Now I'm going to rewrite the left side by breaking apart the fraction into
4:36:59
cosine squared theta over cosine squared theta plus sine squared theta over cosine squared theta. Now cosine squared theta over cosine squared theta is just one. And I can rewrite the next fraction as sine of theta over cosine of theta squared. That's because when I square a fraction, I can just square the numerator and square the denominator. And sine squared theta is shorthand for sine of theta squares. Similarly, for cosine squared theta. Now on
4:37:27
the other side of the equal sign, I can rewrite this fraction as one over cosine theta squared. Again, that's because when I square the fraction, I just get the one squared, which is one divided by the cosine theta squared, which is this. I'm almost done. sine theta over cosine theta is the same thing as tangent theta. And one over cosine theta is the same thing as secant theta. Using the shorthand notation, that says one plus tan squared theta equals sequencer
4:38:07
data, which, after rearranging is exactly the identity that we were looking for. The proof of the third for that green identity is very similar. Once again, I'll start with the identity is to use algebra and to use other identities, like the Pythagorean identity to rewrite one side of the equation till it looks like the other side. The best way to prove the net equation is not an identity is to plug in numbers that break the identity. That is make
4:38:52
the equation not true. Now if you're just trying to decide if an equation has an identity or not, and not worried about proving it, then I recommend plugging in numbers, or graphing the left and right sides to see if those graphs are the same. Recall that an identity is an equation that holds for all values of the variable. This video states and proves three identities called the Pythagorean identities. The first one is the familiar cosine squared
4:39:28
theta plus sine squared theta equals one. The second one says tan squared theta plus one equals secant squared theta. And the third one goes cotangent squared theta plus one equals cosecant squared theta. Let's start by proving that cosine squared theta plus sine squared theta equals one. I'll do this by drawing the unit circle with a right triangle inside it by the definition of sine and cosine, the x&y coordinates of this top point, r cosine
4:39:59
theta and sine theta, the high partners, my triangle is one, since that's the radius of my unit circle. Now the length of the base of my triangle is the same thing as the x coordinate of this point. So that's equal to cosine theta. The height of this triangle is the same thing as the y coordinate of this point. So that's sine theta. Now the Pythagoras theorem for right triangles, says this side length squared plus that side length squared
4:40:23
is equal to the hypothenar squared. So by the Pythagorean theorem, we have that cosine theta squared plus sine theta squared equals one squared, I can rewrite that as cosine squared theta plus sine squared theta equals one, since one squared is one, and cosine squared theta is just a shorthand notation for cosine theta squared. That completes the proof of the first Pythagorean identity, at least in the case, when the angle theta is
4:41:03
in the first quadrant. In the case, when the angle was in a different quadrant, you can use symmetry to argue the same identity holds. But I won't give the details here. To prove the next Pythagorean identity, tan squared theta plus one equals secant squared theta, let's use the first without your an identity, which said that cosine squared theta plus sine squared theta equals one, I'm going to divide both sides of this equation by cosine
4:41:32
squared theta. Now I'm going to rewrite the left side by breaking apart the fraction into cosine squared theta over cosine squared theta plus sine squared theta over cosine squared theta. Now cosine squared theta over cosine squared theta is just one. And I can rewrite the next fraction as sine of theta over cosine of theta squared. That's because when I square a fraction, I can just square the numerator and square the denominator. And sine squared
4:42:08
theta is shorthand for sine of theta squares. Similarly, for cosine squared theta. Now on the other side of the equal sign, I can rewrite this fraction as one over cosine theta squared. Again, that's because when I square the fraction, I just get the one squared, which is one divided by the cosine theta squared, which is this. I'm almost done. sine theta over cosine theta is the same thing as tangent theta. And one over cosine theta is the same thing as secant
4:42:41
theta. Using the shorthand notation, that says one plus tan squared theta equals sequencer data, which, after rearranging is exactly the identity that we were looking for. The proof of the third for that green identity is very similar. Once again, I'll start with the identity cosine squared theta plus sine squared theta equals one, and this time, I'll divide both sides by sine squared theta. I'll break up the fraction on the left. And now I'll rewrite
4:43:07
my fractions as cosine theta over sine theta squared plus one equals one over sine theta squared. cosine over sine can be written As cotangent, and one over sine can be written as cosecant. That gives me the identity that I'm looking for. We've now proved three trig identities. The first one, we proved using the unit circle, and the Pythagorean Theorem. The second and third identities, we proved by using the first identity, and a bit of
4:43:44
algebra. The sum and difference formulas are formulas for computing the sine of a sum of two angles, the cosine of a sum of two angles, the sine of a difference of two angles, and the cosine of a difference of two angles. Please pause the video for a moment to think about this question. Is it true that the sine of A plus B is equal to the sine of A plus the sine of B? No, it's not true. And we can see by an example, if we plug in say, A equals
4:44:22
pi over two and B equals pi, than the sine of pi over two plus pi, is the same thing as a sine of three pi over two, which is negative one. Whereas the sine of pi over two plus the sine of pi is equal to one plus zero, which is one, negative one is not equal to one. So this equation does not hold for all values of a and b. There are a few values of a and b for which it does hold. For example, if a is zero, and B is zero, but it's not
4:44:56
true in general, instead, we need more complicated formulas. It turns out that the sine of the sum of two angles A plus B is given by sine of A cosine of B plus cosine of A, sine of B. The cosine of A plus B is given by cosine A cosine B minus sine A sine Bay. I like to remember these with a song, sine cosine cosine sine cosine cosine minus sine sine. Please feel free to back up the video and sing along with me, I encourage you to memorize the two formulas for the sine of a
4:45:46
cosine squared theta plus sine squared theta equals one, and this time, I'll divide both sides by sine squared theta. I'll break up the fraction on the left. And now I'll rewrite my fractions as cosine theta over sine theta squared plus one equals one over sine theta squared. cosine over sine can be written As cotangent, and one over sine can be written as cosecant. That gives me the identity that I'm looking for. We've now proved three trig
4:46:19
identities. The first one, we proved using the unit circle, and the Pythagorean Theorem. The second and third identities, we proved by using the first identity, and a bit of algebra. The sum and difference formulas are formulas for computing the sine of a sum of two angles, the cosine of a sum of two angles, the sine of a difference of two angles, and the cosine of a difference of two angles. Please pause the video for a moment to think
4:46:57
about this question. Is it true that the sine of A plus B is equal to the sine of A plus the sine of B? No, it's not true. And we can see by an example, if we plug in say, A equals pi over two and B equals pi, than the sine of pi over two plus pi, is the same thing as a sine of three pi over two, which is negative one. Whereas the sine of pi over two plus the sine of pi is equal to one plus zero, which is one, negative one is not equal to
4:47:39
one. So this equation does not hold for all values of a and b. There are a few values of a and b for which it does hold. For example, if a is zero, and B is zero, but it's not true in general, instead, we need more complicated formulas. It turns out that the sine of the sum of two angles A plus B is given by sine of A cosine of B plus cosine of A, sine of B. The cosine of A plus B is given by cosine A cosine B minus sine A sine Bay. I like to
4:48:17
remember these with a song, sine cosine cosine sine cosine cosine minus sine sine. Please feel free to back up the video and sing along with me, I encourage you to memorize the two formulas for the sine of a sum of angles and the cosine of a sum of angles. Once you do, it's easy to figure out the sine sum of angles and the cosine of a sum of angles. Once you do, it's easy to figure out the sine and cosine of a difference of two angles. One way to do this is to think of sine of
4:49:52
and cosine of a difference of two angles. One way to do this is to think of sine of A minus B as sine of A plus negative B. And then use the angle sum formula. So this works out to sine cosine plus cosine, sine. And now, if I use the fact that cosine is even, I know that cosine of negative B is cosine of B. And since sine is odd, sine of negative b is negative sine of B. So I can rewrite this as sine of A cosine of B minus cosine
4:50:39
of A sine of B. Notice that this new formula for the difference is the same as the formula for the sum is just that plus sign turned into a minus sign. We can do the same trick for cosine of A minus B, that's cosine of A plus minus b, which is cosine A cosine minus b minus sine of A sine of negative B. Again, using even an odd properties, this gives us cosine A cosine B plus sine A sine B. Once again, the formula for the difference is almost
4:51:19
exactly like the for the song, just that minus sign has switched to a plus sign. Now let's use the angle sum formula to find the exact value for the sign of 105 degrees. Now, 105 degrees is not a special angle on the unit circle, but I can write it as the sum of two special angles. I can write it as 60 degrees plus 45 degrees. Therefore, the sine of 105 degrees is the sine of 60 plus 45. And now by the angle some formula,
4:52:04
this is sine, cosine, cosine, sine. And I for my Unit Circle, I can figure out that sine of 60 degrees is root three over two cosine of 45 degrees root two over two, cosine of 60 degrees is one half, and sine of 45 degrees is root two over two. So this simplifies to root six plus root two over four. For our last example, let's find the cosine of v plus W, given the values of cosine v and cosine W, and the fact that v and w are angles in
4:52:49
the first quadrant. Remember, to compute the cosine of a sum, we can't just add together the two cosines. That wouldn't even make sense in this case, because adding point nine and point seven would give something bigger than one and the cosine of something's never bigger than one. Instead, we have to use the angle sum formula for cosine. So that goes cosine of v plus w equals cosine, cosine, minus sine, sine. Now, I've already know the cosine of
4:53:17
v and the cosine of W, so I could just plug those in. But I have to figure out the sine of v and the sine of W from the given information. And one way to do that is to draw right triangles. So here, I'm going to draw a right triangle with angle V, A minus B as sine of A plus negative B. And then use the angle sum formula. So this works out to sine cosine plus cosine, sine. And now, if I use the fact that cosine is even,
4:53:51
I know that cosine of negative B is cosine of B. And since sine is odd, sine of negative b is negative sine of B. So I can rewrite this as sine of A cosine of B minus cosine of A sine of B. Notice that this new formula for the difference is the same as the formula for the sum is just that plus sign turned into a minus sign. We can do the same trick for cosine of A minus B, that's cosine of A plus minus b, which is cosine A cosine minus
4:55:49
b minus sine of A sine of negative B. Again, using even an odd properties, this gives us cosine A cosine B plus sine A sine B. Once again, the formula for the difference is almost exactly like the for the song, just that minus sign has switched to a plus sign. Now let's use the angle sum formula to find the exact value for the sign of 105 degrees. Now, 105 degrees is not a special angle on the unit circle, but I can write it as the sum of two
4:56:34
special angles. I can write it as 60 degrees plus 45 degrees. Therefore, the sine of 105 degrees is the sine of 60 plus 45. And now by the angle some formula, this is sine, cosine, cosine, sine. And I for my Unit Circle, I can figure out that sine of 60 degrees is root three over two cosine of 45 degrees root two over two, cosine of 60 degrees is one half, and sine of 45 degrees is root two over two. So this simplifies to root six plus root
4:57:08
two over four. For our last example, let's find the cosine of v plus W, given the values of cosine v and cosine W, and the fact that v and w are angles in the first quadrant. Remember, to compute the cosine of a sum, we can't just add together the two cosines. That wouldn't even make sense in this case, because adding point nine and point seven would give something bigger than one and the cosine of something's never bigger than one.
4:57:35
Instead, we have to use the angle sum formula for cosine. So that goes cosine of v plus w equals cosine, cosine, minus sine, sine. Now, I've already know the cosine of v and the cosine of W, so I could just plug those in. But I have to figure out the sine of v and the sine of W from the given information. And one way to do that is to draw right triangles. So here, I'm going to draw a right triangle with angle V,
4:58:13
and another right triangle with angle W. Since I know that the cosine of V is point nine, I can think of that as nine over 10. and another right triangle with angle W. Since I know that the cosine of V is point nine, I can think of that as nine over 10. And I can think of that as adjacent over hypotenuse in my right triangle. So I'll decorate my triangles adjacent side with the number nine and the hypotenuse with 10. Similarly, since
4:58:49
I know that the cosine of W is point seven, which is seven tenths, I can put a seven on this adjacent side, and a 10 on this iPod news. Now, the Pythagorean Theorem, lets me compute the length of the unlabeled side. So this one is going to be the square root of 10 squared minus nine squared, that's going to be the square root of 19. And here I have the square root of 10 squared minus seven squared. So that's the square root of 51.
4:59:33
I can now find the sign of V as the opposite over the hi partners. So that's the square root of 19 over 10. And the sign of W will be the square root of 51 over 10. Because we're assuming v and w are in the first quadrant, we know the values of sign need to be positive, so we don't need to Jimmy around with positive or negative signs in our answers, we can just leave them as is. And I can think of that as adjacent over hypotenuse in my right triangle. So I'll decorate my
5:00:42
triangles adjacent side with the number nine and the hypotenuse with 10. Similarly, since I know that the cosine of W is point seven, which is seven tenths, I can put a seven on this adjacent side, and a 10 on this iPod news. Now, the Pythagorean Theorem, lets me compute the length of the unlabeled side. So this one is going to be the square root of 10 squared minus nine squared, that's going to be the square root of 19. And here I have
5:01:32
the square root of 10 squared minus seven squared. So that's the square root of 51. I can now find the sign of V as the opposite over the hi partners. So that's the square root of 19 over 10. And the sign of W will be the square root of 51 over 10. Because we're assuming v and w are in the first quadrant, we know the values of sign need to be positive, so we don't need to Jimmy around with positive or negative signs in our answers, we can just leave them as is.
5:02:14
Now we're ready to plug into our formula. So we have that cosine of v plus w is equal to point nine times point seven minus the square root of 19 over 10, times the square root of 51 over 10. Now we're ready to plug into our formula. So we have that cosine of v plus w is equal to point nine times point seven minus the square root of 19 over 10, times the square root of 51 over 10. using a calculator, this works out to a decimal approximation of 0.3187. This video gave the
5:02:49
angle sum and difference formulas and use them to compute some values. To see a proof for why the sum formulas hold, please watch my other video. This video gives formulas for sine of two theta and cosine of two theta. Please pause the video for a moment and see if you think this equation sine of two theta equals two sine theta is true or false. Remember that true means always true for all values of theta were false means sometimes they're
5:03:20
always false. This equation is false, because it's not true for all values of theta. One way to see this is graphically, if I graph y equals sine of two theta, that's like the graph of sine theta, squished in horizontally by a factor of one half. On the other hand, if I graph y equals two sine beta, that's like the graph of sine theta stretched vertically by a factor of two. These two graphs are not the same. So instead, we need a more complicated
5:03:58
formula for sine of two theta. And that formula is sine of two theta is two sine theta, cosine theta. It's not hard to see why that formula works based on the angle some formula. Recall that sine of A plus B is equal to sine A cosine B plus cosine A sign Therefore, sine of two theta, which is sine of theta plus theta is going to be sine theta, cosine theta, plus cosine theta sine theta. simply plugging in theta for a and theta for B, in this angle,
5:04:48
some formula, will sine theta cosine theta is the same thing as cosine theta sine theta. So I can rewrite this as twice sine theta cosine theta. That gives me this formula. There's also a formula for cosine of two theta. And that formula is cosine squared theta minus sine squared theta. Again, we can use the angle sum formula to see where this comes from. cosine of A plus B is equal to cosine of A cosine of B minus sine A, sine B. So
5:05:27
if we want cosine of two theta, that's just cosine of theta plus theta, which is cosine theta, cosine theta, minus sine theta, sine theta by plugging in beta for a and b, this can be rewritten as cosine squared theta minus sine squared theta, which is exactly the formula above. Now there are a couple other formulas for cosine of two theta that are also popular. One of them is one minus two sine squared theta. And the other one is cosine of two
5:06:06
theta is two cosine squared theta minus one, you can get each of these two formulas from the original one using the Pythagorean identity. We know that cosine squared theta plus sine squared theta is one. So cosine squared theta is one minus sine squared theta. If I plug that into my original formula, which I've copied here, so I'm plugging in, instead of cosine squared, I'm gonna write one minus sine squared theta, I still have a nother
5:06:39
minus sine squared theta. So that's the same thing as one minus twice science grid data, which is exactly what I'm looking for. Similarly, I can use the Pythagorean identity to write sine squared theta as one minus cosine squared theta. Again, I'll take this equation and copy it below. But this time, I'm going to plug in for sine squared right here. So that gives me cosine of two theta is cosine squared theta minus the quantity one minus cosine
5:07:18
squared theta. That simplifies to two cosine squared theta minus one after distributing the negative sign and combining like terms. using a calculator, this works out to a decimal approximation of 0.3187. This video gave the angle sum and difference formulas and use them to compute some values. To see a proof for why the sum formulas hold, please watch my other video. This video gives formulas for sine of two theta and cosine of two theta. Please pause the video for a moment and see
5:08:04
if you think this equation sine of two theta equals two sine theta is true or false. Remember that true means always true for all values of theta were false means sometimes they're always false. This equation is false, because it's not true for all values of theta. One way to see this is graphically, if I graph y equals sine of two theta, that's like the graph of sine theta, squished in horizontally by a factor of one half. On the other hand,
5:08:53
if I graph y equals two sine beta, that's like the graph of sine theta stretched vertically by a factor of two. These two graphs are not the same. So instead, we need a more complicated formula for sine of two theta. And that formula is sine of two theta is two sine theta, cosine theta. It's not hard to see why that formula works based on the angle some formula. Recall that sine of A plus B is equal to sine A cosine B plus cosine A sign Therefore, sine of two
5:09:53
theta, which is sine of theta plus theta is going to be sine theta, cosine theta, plus cosine theta sine theta. simply plugging in theta for a and theta for B, in this angle, some formula, will sine theta cosine theta is the same thing as cosine theta sine theta. So I can rewrite this as twice sine theta cosine theta. That gives me this formula. There's also a formula for cosine of two theta. And that formula is cosine squared theta minus
5:10:33
sine squared theta. Again, we can use the angle sum formula to see where this comes from. cosine of A plus B is equal to cosine of A cosine of B minus sine A, sine B. So if we want cosine of two theta, that's just cosine of theta plus theta, which is cosine theta, cosine theta, minus sine theta, sine theta by plugging in beta for a and b, this can be rewritten as cosine squared theta minus sine squared theta, which is exactly the formula
5:11:13
above. Now there are a couple other formulas for cosine of two theta that are also popular. One of them is one minus two sine squared theta. And the other one is cosine of two theta is two cosine squared theta minus one, you can get each of these two formulas from the original one using the Pythagorean identity. We know that cosine squared theta plus sine squared theta is one. So cosine squared theta is one minus sine squared theta. If I plug
5:11:36
that into my original formula, which I've copied here, so I'm plugging in, instead of cosine squared, I'm gonna write one minus sine squared theta, I still have a nother minus sine squared theta. So that's the same thing as one minus twice science grid data, which is exactly what I'm looking for. Similarly, I can use the Pythagorean identity to write sine squared theta as one minus cosine squared theta. Again, I'll take this equation and
5:11:55
copy it below. But this time, I'm going to plug in for sine squared right here. So that gives me cosine of two theta is cosine squared theta minus the quantity one minus cosine squared theta. That simplifies to two cosine squared theta minus one after distributing the negative sign and combining like terms. So I have one double angle formula for sine of two theta. And I have three versions of the double angle formula for cosine of two theta.
5:12:30
So I have one double angle formula for sine of two theta. And I have three versions of the double angle formula for cosine of two theta. Now let's use these formulas in some examples. Let's find the cosine of two theta. If we know that cosine theta is negative one over root 10, and theta terminates in quadrant three, we have a choice of three formulas for cosine of two theta, I'm going to choose the second one, because it only involves cosine if they had on the right side. And they already
5:13:20
know my value for cosine theta. Of course, I could use one of the other ones, Now let's use these formulas in some examples. Let's find the cosine of two theta. If we know that cosine theta is negative one over root 10, and theta terminates in quadrant three, we have a choice of three formulas for cosine of two theta, I'm going to choose the second one, because it only involves cosine if they had on the right side. And they already
5:13:50
know my value for cosine theta. Of course, I could use one of the other ones, but then I'd have to work out the value of sine theta. So plugging in, I get cosine of two theta is twice negative one over root n squared minus one, which simplifies to two tenths minus one or negative eight tenths, negative four fifths. Finally, let's solve the equation two cosine x plus sine of 2x equals zero. What makes this equation tricky
5:14:16
is that one of the trig functions has the argument of just x, but the other tree function has the argument of 2x. So I want to use my double angle formula to rewrite sine of 2x. I'll copy down the two cosine x, and now sine of 2x is equal to two sine x cosine x. At this point, I see a way to factor my equation, I can factor out a two cosine x from both of these two terms. That gives me one plus sine x and the product there is equal to zero.
5:15:07
That means that either two cosine x is equal to zero One plus sine x is equal to zero, that simplifies to cosine x equals zero, or sine x is negative one. Using my unit circle, I see that cosine of x is zero at pi over two, and three pi over two, while sine of x is negative one at three pi over two, there's some redundancy here, but my solution set is going to be pi over two plus multiples of two pi, and three pi over two plus multiples
5:15:55
of two pi. This video proved the double angle formulas, sine of two theta is two sine theta cosine theta. and cosine of two theta is cosine squared theta minus sine squared theta. It also proved to alternate versions of the equation for cosine of two theta. This video introduces higher order derivatives and notation. We've seen that f prime of x denotes the derivative of the function f of x, but f prime of x is also itself a function. So we can take its
5:16:34
derivative, that would be f prime prime of x, which is usually written instead as f double prime of x. This is called the second derivative of f. And it means the derivative of the derivative, we can also talk about the third derivative, f triple prime of x, which might sometimes be written f to the three of x if you get tired of writing all those primes, and we can talk about the nth derivative f, parentheses n of x, the parentheses here are important
5:17:09
to show that it's the nth derivative. The second, third and nth derivative are referred to as higher order but then I'd have to work out the value of sine theta. So plugging in, I get cosine of two theta is twice negative one over root n squared minus one, which simplifies to two tenths minus one or negative eight tenths, negative four fifths. Finally, let's solve the equation two cosine x plus sine of 2x equals zero. What makes this equation tricky
5:17:36
is that one of the trig functions has the argument of just x, but the other tree function has the argument of 2x. So I want to use my double angle formula to rewrite sine of 2x. I'll copy down the two cosine x, and now sine of 2x is equal to two sine x cosine x. At this point, I see a way to factor my equation, I can factor out a two cosine x from both of these two terms. That gives me one plus sine x and the product there is equal to zero.
5:18:25
That means that either two cosine x is equal to zero One plus sine x is equal to zero, that simplifies to cosine x equals zero, or sine x is negative one. Using my unit circle, I see that cosine of x is zero at pi over two, and three pi over two, while sine of x is negative one at three pi over two, there's some redundancy here, but my solution set is going to be pi over two plus multiples of two pi, and three pi over two plus multiples
5:19:14
of two pi. This video proved the double angle formulas, sine of two theta is two sine theta cosine theta. and cosine of two theta is cosine squared theta minus sine squared theta. It also proved to alternate versions of the equation for cosine of two theta. This video introduces higher order derivatives and notation. We've seen that f prime of x denotes the derivative of the function f of x, but f prime of x is also itself a function. So we can take its
5:19:47
derivative, that would be f prime prime of x, which is usually written instead as f double prime of x. This is called the second derivative of f. And it means the derivative of the derivative, we can also talk about the third derivative, f triple prime of x, which might sometimes be written f to the three of x if you get tired of writing all those primes, and we can talk about the nth derivative f, parentheses n of x, the parentheses here are important
5:20:24
to show that it's the nth derivative. The second, third and nth derivative are referred to as higher order derivatives. derivatives. There are many alternative notations for derivatives stemming from their tangled and contentious history in the 1600s. There are a few different notations for functions themselves. Most often we write a function of something like f of x, but we can also use the variable y to refer
5:20:50
to the output of a function. When we're looking at the first derivative. We've been using the notation f prime of x. But you might also see y prime, this means the same thing. Another notation is df dx, is known as lug nuts denotation. After five minutes, you might see something like dy dx of f of x. And you might see dy dx, another version of Lebanon's notation. Sometimes you'll also see a capital D used to refer to the derivative. If we're looking
5:21:16
at the second derivative, we've seen that in Haitian f double prime of x, y double prime is a similar notation, or we might write dy dx of df There are many alternative notations for derivatives stemming from their tangled and contentious history in the 1600s. There are a few different notations for functions themselves. Most often we write a function of something like f of x, but we can also use the variable y to refer
5:21:34
to the output of a function. When we're looking at the first derivative. We've been using the notation f prime of x. But you might also see y prime, this means the same thing. Another notation is df dx, is known as lug nuts denotation. After five minutes, you might see something like dy dx of f of x. And you might see dy dx, another version of Lebanon's notation. Sometimes you'll also see a capital D used to refer to the derivative. If we're looking
5:22:13
at the second derivative, we've seen that in Haitian f double prime of x, y double prime is a similar notation, or we might write dy dx of df dx. dx. And the shorthand for that is d squared f dx squared. Similarly, we might write d squared y dx squared using y in the place of F for the function. And the shorthand for that is d squared f dx squared. Similarly, we might write d squared y dx squared using y in the place of F for the function.
5:22:53
There's similar notations for third derivative, I'll jump ahead to nth derivatives. So that would be F to the n of x, or y to the n, d to the n of f dx to the n, or D to the n of y dx to the N. When using live minutes notation, we want to emphasize that we're evaluating our derivative at a particular value of x, we might write something like at x equals three, or at x equals a, using a vertical line. For better or for worse, you'll need to become familiar with all of these alternative notations.
5:24:13
That's all for this video on higher order derivatives and notation. This video is about the derivative of e to the x, one of my favorite functions ever simply because it has such a great derivative. As you may recall, he is an irrational number whose decimal approximation is something like 2.718 looks like it's repeating. But then it keeps going on forever, never repeating never terminating. That's all for this video on higher order derivatives and notation. This video is about
5:24:48
the derivative of e to the x, one of my favorite functions ever simply because it has such a great derivative. As you may recall, he is an irrational number whose decimal approximation is something like 2.718 looks like it's repeating. But then it keeps going on forever, never repeating never terminating. Its value notice is somewhere in between two and three. Here is a graph of y equals e to the x. Its value notice is somewhere in between two and three. Here is a graph of y equals e to the x.
5:25:29
It says exponential function increase thing looks a lot like two to the x or three to the x, not only is the graph of e to the x increasing, but it's increasing more and more rapidly. So for negative values of x, the slope of this graph is positive but very close to zero. Over here, when x equals zero, that slope is looks like approximately a slope of one, we'll see that it is in fact exactly one. And as the x values increase, this tangent
5:26:09
lines get steeper and steeper. I'm going to state without proof, three really useful facts about E. First, if you take the limit, as n goes to infinity of one plus one over N raised to the nth power, that limit exists and equals E, you might have seen something like that when you took precalculus, and looked at compound interest compounded over smaller and smaller time periods. But even if you haven't seen it before, it's a really important
5:26:35
fact worth memorizing, you'll see it again later in the class. A second important formula is that the limit as h goes to zero of e to the H minus one over h equals one. Now this expression here on the left may remind you of a derivative, in fact, I can rewrite it as the limit as h goes to zero of e to the zero plus h minus e to the zero since either the zero is one over h, that's equal to one. And this expression right here on the left
5:27:11
is just the derivative of e to the x at x equals zero, according to the limit definition of derivative. So this fact, is really saying that the derivative of e to the x at x equals zero, that derivative is equal to one. So for the third fat, It says exponential function increase thing looks a lot like two to the x or three to the x, not only is the graph of e to the x increasing, but it's increasing more and more
5:27:35
rapidly. So for negative values of x, the slope of this graph is positive but very close to zero. Over here, when x equals zero, that slope is looks like approximately a slope of one, we'll see that it is in fact exactly one. And as the x values increase, this tangent lines get steeper and steeper. I'm going to state without proof, three really useful facts about E. First, if you take the limit, as n goes to infinity of one plus one over N
5:28:10
raised to the nth power, that limit exists and equals E, you might have seen something like that when you took precalculus, and looked at compound interest compounded over smaller and smaller time periods. But even if you haven't seen it before, it's a really important fact worth memorizing, you'll see it again later in the class. A second important formula is that the limit as h goes to zero of e to the H minus one over h equals one. Now this
5:28:38
expression here on the left may remind you of a derivative, in fact, I can rewrite it as the limit as h goes to zero of e to the zero plus h minus e to the zero since either the zero is one over h, that's equal to one. And this expression right here on the left is just the derivative of e to the x at x equals zero, according to the limit definition of derivative. So this fact, is really saying that the derivative of e to the x at x equals
5:29:18
zero, that derivative is equal to one. So for the third fat, the third fact, the third fact, it talks about the derivative of e to the x in general. And that third fact is that the derivative of the function, either the X is the function, e to the x, e to the x is its own derivative. So this is a generalized version of the second fact. Because the second fact is saying that the derivative at x equals zero is one, well, one is just the same thing
5:30:07
as e to the zero. So it saying the drought of either the exit at x equals zero is equal to zero. And in general, the derivative of e to the x at any x is just e to the x. Now, fact, one is frequently taken as the definition of E. Sometimes fact two instead is taken as a definition of he, since he is the unique number with this, this property, the unique number, you can plug in here and get this limit equal one, it's possible to prove that
5:30:34
fact one implies fact two and vice versa. But I won't do that here. It's also possible to prove that fact two implies fact three about the derivative in general. And that's pretty straightforward from the definition of derivative. So I will show you that argument. So let's start out assuming fact to and try to prove fact three using the definition of direct by the definition of derivative. The derivative of e to the x is the limit as h
5:31:03
goes to zero of e to the x plus h minus e to the x over h. If I factor out an E to the X, from both terms on the numerator, I get the limit of e to the x times e to the H minus one over h. Notice that e to the x times e to the H is in the x plus h by the exponent rules. Now, either the X has nothing to do with H, so it's just a constant as far as H is concerned. And I can pull it all the way out of the limit sign and rewrite this
5:31:49
limit. Now by factor two, which I'm assuming this limit here is just one, which means that my derivative it talks about the derivative of e to the x in general. And that third fact is that the derivative of the function, either the X is the function, e to the x, e to the x is its own derivative. So this is a generalized version of the second fact. Because the second fact is saying that the derivative at x equals zero is one, well, one is just the same thing
5:32:24
as e to the zero. So it saying the drought of either the exit at x equals zero is equal to zero. And in general, the derivative of e to the x at any x is just e to the x. Now, fact, one is frequently taken as the definition of E. Sometimes fact two instead is taken as a definition of he, since he is the unique number with this, this property, the unique number, you can plug in here and get this limit equal one, it's possible to prove that
5:33:02
fact one implies fact two and vice versa. But I won't do that here. It's also possible to prove that fact two implies fact three about the derivative in general. And that's pretty straightforward from the definition of derivative. So I will show you that argument. So let's start out assuming fact to and try to prove fact three using the definition of direct by the definition of derivative. The derivative of e to the x is the limit as h
5:33:47
goes to zero of e to the x plus h minus e to the x over h. If I factor out an E to the X, from both terms on the numerator, I get the limit of e to the x times e to the H minus one over h. Notice that e to the x times e to the H is in the x plus h by the exponent rules. Now, either the X has nothing to do with H, so it's just a constant as far as H is concerned. And I can pull it all the way out of the limit sign and rewrite this
5:34:34
limit. Now by factor two, which I'm assuming this limit here is just one, which means that my derivative is e to the x, just like I wanted to show. Here's a slightly tricky example, asking you to compute the derivative of a function that involves lots of ears and X's is e to the x, just like I wanted to show. Here's a slightly tricky example, asking you to compute the derivative of a function that involves lots of ears and X's
5:35:13
combined in lots of different ways. You'll need to use not only the rule for the derivative of e to the x that we just talked about, but also the power rule and other rules. derivatives that we've talked about combined in lots of different ways. You'll need to use not only the rule for the derivative of e to the x that we just talked about, but also the power rule and other rules. derivatives that we've talked about earlier. earlier.
5:35:54
So please pause the video and try to compute this derivative yourself paying careful attention to what's a variable and what's the constant. So please pause the video and try to compute this derivative yourself paying careful attention to what's a variable and what's the constant. Okay, so we're taking the derivative here with respect to x, that's our variable. And I'm taking the derivative of this entire expression, which I can split up as a sum of derivatives.
5:36:27
For the first term, I can just use the power rule, E is a constant coefficient, so I just need to take down the exponent of two multiplied on the front, times x to the One Power. Now for the second part, Okay, so we're taking the derivative here with respect to x, that's our variable. And I'm taking the derivative of this entire expression, which I can split up as a sum of derivatives. For the first term, I can just use the power rule, E is a constant coefficient, so I just
5:37:15
need to take down the exponent of two multiplied on the front, times x to the One Power. Now for the second part, here, I do have my either the x function multiplied by two, so its derivative is just two times the derivative of e to the x, which is either the X here, I do have my either the x function multiplied by two, so its derivative is just two times the derivative of e to the x, which is either the X
5:37:51
for my third part, I have just x times a constant e squared. So the derivative of x is one times that constant. And so I just get e squared. Finally, to take the derivative of x to the power of E squared, I can use the power rule because my variable is in the base, and I have a constant e squared in my exponent. So using the power rule, I bring down the E squared times that by x and subtract one from the exponent. This video states the happy fact that the derivative of e to the x is just e to
5:39:30
the x. This video prove some of the rules for taking derivatives. the x. This video prove some of the rules for taking derivatives. First, the constant row, it makes sense that the derivative of a constant real number has to be zero, because the slope of a horizontal line First, the constant row, it makes sense that the derivative of a constant real number has to be zero, because the slope of a horizontal line is zero. is zero.
5:40:02
But we can also prove this fact, using the limit definition of derivative. The derivative of any function is the limit as h goes to zero of the function of x plus h minus the function at x divided by H. Well, here, our function is just a constant. So we're taking the limit as h goes to zero of the constant minus the constant divided by h, which is just the limit as h goes to zero of zero over h, which is just the limit of zero, which
5:40:43
is zero. Intuitively, it also makes sense that the derivative of the function y equals x is got to be one, because the graph of y equals x is a straight line with slope one. But again, we can prove this using the limit definition of derivative. So the derivative of x is the limit as h goes to zero of x plus h minus x over h. Well, that simplifies to the limit of h over h, since the Xs can't So in other words, the limit as h goes to zero of one, which is one as wanted
5:41:28
But we can also prove this fact, using the limit definition of derivative. The derivative of any function is the limit as h goes to zero of the function of x plus h minus the function at x divided by H. Well, here, our function is just a constant. So we're taking the limit as h goes to zero of the constant minus the constant divided by h, which is just the limit as h goes to zero of zero over h, which is
5:44:26
just the limit of zero, which is zero. Intuitively, it also makes sense that the derivative of the function y equals x is got to be one, because the graph of y equals x is a straight line with slope one. But again, we can prove this using the limit definition of derivative. So the derivative of x is the limit as h goes to zero of x plus h minus x over h. Well, that simplifies to the limit of h over h, since the Xs can't So in other words, the
5:44:58
limit as h goes to zero of one, which is one as wanted x, x, h to the n minus one, and then finally, a term of h to the N. That's the binomial expansion of x plus h to the n. Now, we still have to subtract the x to the n that we had up here, and we still have to divide this whole thing by H. Okay, that's looking kind of horribly complicated. h to the n minus one, and then finally, a term of h to the N. That's the binomial expansion
5:45:53
of x plus h to the n. Now, we still have to subtract the x to the n that we had up here, and we still have to divide this whole thing by H. Okay, that's looking kind of horribly complicated. But But notice that the X to the ends can So notice that all of the remaining terms have an H in them. So if we factor out that H, we get an x to the n minus one, plus a bunch of other terms. And canceling the H's, notice that the X to the ends can So notice that all of the remaining terms have an H
5:46:42
in them. So if we factor out that H, we get an x to the n minus one, plus a bunch of other terms. And canceling the H's, we get we get one term that doesn't have any ages in it, and another bunch of terms that all have H's in them, as h goes to zero, all these other terms drop out because they go to zero. And what we're left with is simply n times x to the n minus one, which is exactly what we want for the power role. I think that's a pretty good proof if you're comfortable with
5:47:20
the binomial formula. But if you haven't seen the binomial formula before, that might leave you feeling a little cold. So I'm going to offer you another proof using the other form of the limit definition of derivative. So let me clear some space here. And I'll start over using this definition, f prime at a is the limit as x goes to a of our function evaluated at x. So that's x to the n minus our function evaluated at a, that's a to the n over x minus
5:47:52
a. Again, I'm going to need to rewrite things in order to evaluate this limit, since it's currently in a zero over zero and determinant form. So I'm going to rewrite the top by factoring out a copy of x minus a, which gives me x to the n minus one plus x to the n minus two A plus x to the n minus three A squared, you see the pattern here. And I keep going until I get to x a to the n minus two and finally a to the n minus one, that's still over x
5:48:42
minus a, you can verify this factoring formula, simply by multiplying out and checking that you In fact, do get x to the n minus a to the n, after all your intermediate terms cancel. Now that I've factored, I can cancel my x minus a, and simply evaluate my limit by plugging in x equal to A to get a to the n minus one plus a to the n minus two A plus, and so on each of these terms is equal to a to the n minus one. And there are a total of n terms,
5:49:17
since we got them from the terms above that started with x to the n minus one and did with x to the zero. So that's n terms. So that means we've got a final sum of n times a to the n minus one for a derivative f prime of A, which is exactly what we wanted to show. Next, I'll prove the constant multiple rule that says that if c is a real number constant, and f is a differentiable function, then the derivative of a constant times f is just the
5:49:50
constant times the derivative of f. Starting with the limit definition of derivative, I have that the derivative of C times f of x is the limit as h goes to zero of C times f of x plus h minus c times f of x over h. Now if I factor out the constant C, from both of these terms, and actually I can pull it all the way out of the limit side, since the constant has nothing to do with h. So now I get that was equal to the constant times
5:50:16
the limit as h goes to zero of f of x plus h minus f of x over h, which is just a constant times the derivative of f, which is what we wanted to prove. one term that doesn't have any ages in it, and another bunch of terms that all have H's in them, as h goes to zero, all these other terms drop out because they go to zero. And what we're left with is simply n times x to the n minus one, which is exactly what we want for the power role. I think that's a pretty good proof if you're comfortable with
5:51:02
the binomial formula. But if you haven't seen the binomial formula before, that might leave you feeling a little cold. So I'm going to offer you another proof using the other form of the limit definition of derivative. So let me clear some space here. And I'll start over using this definition, f prime at a is the limit as x goes to a of our function evaluated at x. So that's x to the n minus our function evaluated at a, that's a to the n over x minus
5:51:36
a. Again, I'm going to need to rewrite things in order to evaluate this limit, since it's currently in a zero over zero and determinant form. So I'm going to rewrite the top by factoring out a copy of x minus a, which gives me x to the n minus one plus x to the n minus two A plus x to the n minus three A squared, you see the pattern here. And I keep going until I get to x a to the n minus two and finally a to the n minus one, that's still over x
5:52:02
minus a, you can verify this factoring formula, simply by multiplying out and checking that you In fact, do get x to the n minus a to the n, after all your intermediate terms cancel. Now that I've factored, I can cancel my x minus a, and simply evaluate my limit by plugging in x equal to A to get a to the n minus one plus a to the n minus two A plus, and so on each of these terms is equal to a to the n minus one. And there are a total of n terms,
5:52:35
since we got them from the terms above that started with x to the n minus one and did with x to the zero. So that's n terms. So that means we've got a final sum of n times a to the n minus one for a derivative f prime of A, which is exactly what we wanted to show. Next, I'll prove the constant multiple rule that says that if c is a real number constant, and f is a differentiable function, then the derivative of a constant times f is just the
5:53:17
constant times the derivative of f. Starting with the limit definition of derivative, I have that the derivative of C times f of x is the limit as h goes to zero of C times f of x plus h minus c times f of x over h. Now if I factor out the constant C, from both of these terms, and actually I can pull it all the way out of the limit side, since the constant has nothing to do with h. So now I get that was equal to the constant times
5:54:04
the limit as h goes to zero of f of x plus h minus f of x over h, which is just a constant times the derivative of f, which is what we wanted to prove. The difference rule can be proved just like the sum rule by writing out the definition of derivative and regrouping terms. Or we could use kind of a sneaky shortcut, and put together two of our previous roles. So if we think of f of x minus g of x as being f of x plus minus one times g of x, then we can use the sum rule to rewrite this as a
5:54:38
sum of derivatives, and then use the constant multiplier rule to pull the constant of negative one out, and then we have exactly what we wanted to prove. So in this video, we gave the proof of the content multiple rule, the summon difference rules, and a proof of the power rule when n is a positive integer. This video gives rules for calculating derivatives of functions that are products, or quotients of other functions. In this video, you'll
5:55:06
find statements of the product rule and the quotient rule, and some examples. But no proofs, proofs are in a separate video. Before we start, let's recall the Psalm and the difference rules. If f and g are differentiable functions, then the derivative of the psalm f of x plus g of x is just the sum of the derivatives. And a similar statement holds for differences. The derivative of the difference is the difference of the derivatives. So does the same sort
5:55:48
of rule hold for products of functions, in other words, is the derivative of the product equal to the product of the derivatives? Let's look at a simple example. To find out. For example, if f of x is x, and g of x is x squared, then if we take the derivative of the product, x times x squared, well, that's just the derivative of x cubed. We know how to do that with the power rule. So it's 3x squared. On the other hand, if we look at the product of the derivatives,
5:56:38
we get one times 2x or just 2x. And these two things are not equal. So unfortunately, the answer is no such a simple product rule does not hold. But not lose hope. There is a product rule, it's just a little more complicated than the sum and difference rule. The product rule says that if f and g are differentiable functions, then the derivative of the product f of x times g of x is equal to f of x times the derivative
5:57:21
of g of x plus the derivative of f of x times g of x. In other words, to take the derivative of a product, we take the first function times the derivative of the second, plus the derivative of the first times the second. Let's use this in an example. To take the derivative of the square root of t times e to the t, we have to take the first function squared of t times the derivative of the second function, plus the derivative of the first function
5:57:58
The difference rule can be proved just like the sum rule by writing out the definition of derivative and regrouping terms. Or we could use kind of a sneaky shortcut, and put together two of our previous roles. So if we think of f of x minus g of x as being f of x plus minus one times g of x, then we can use the sum rule to rewrite this as a sum of derivatives, and then use the constant multiplier rule to pull the constant of negative
5:58:35
one out, and then we have exactly what we wanted to prove. So in this video, we gave the proof of the content multiple rule, the summon difference rules, and a proof of the power rule when n is a positive integer. This video gives rules for calculating derivatives of functions that are products, or quotients of other functions. In this video, you'll find statements of the product rule and the quotient rule, and some examples. But no proofs,
5:59:04
proofs are in a separate video. Before we start, let's recall the Psalm and the difference rules. If f and g are differentiable functions, then the derivative of the psalm f of x plus g of x is just the sum of the derivatives. And a similar statement holds for differences. The derivative of the difference is the difference of the derivatives. So does the same sort of rule hold for products of functions, in other words, is the derivative of the product
5:59:38
equal to the product of the derivatives? Let's look at a simple example. To find out. For example, if f of x is x, and g of x is x squared, then if we take the derivative of the product, x times x squared, well, that's just the derivative of x cubed. We know how to do that with the power rule. So it's 3x squared. On the other hand, if we look at the product of the derivatives, we get one times 2x or just 2x. And these two things are not equal. So unfortunately,
6:00:33
the answer is no such a simple product rule does not hold. But not lose hope. There is a product rule, it's just a little more complicated than the sum and difference rule. The product rule says that if f and g are differentiable functions, then the derivative of the product f of x times g of x is equal to f of x times the derivative of g of x plus the derivative of f of x times g of x. In other words, to take the derivative of a product, we take
6:01:06
the first function times the derivative of the second, plus the derivative of the first times the second. Let's use this in an example. To take the derivative of the square root of t times e to the t, we have to take the first function squared of t times the derivative of the second function, plus the derivative of the first function times the second function. So that's the square root of t times the derivative of e to the t is just e to the t.
6:01:38
times the second function. So that's the square root of t times the derivative of e to the t is just e to the t. And to find the derivative of the square root of t, it's going to be easier to write it in exponential form. Now we can just use the power roll. Bring down the one half t to the one half minus one is negative one half and I found the derivative. I'm going to clean this up a little bit, and I'm done. quotient rule says that the derivative of a quotient of two functions
6:02:20
And to find the derivative of the square root of t, it's going to be easier to write it in exponential form. Now we can just use the power roll. Bring down the one half t to the one half minus one is negative one half and I found the derivative. I'm going to clean this up a little bit, and I'm done. quotient rule says that the derivative of a quotient of two functions is given By this quotient on the denominator, we have the denominator function g of x squared.
6:02:52
And on the numerator, we have g of x times the derivative is given By this quotient on the denominator, we have the denominator function g of x squared. And on the numerator, we have g of x times the derivative of f of x of f of x minus f of x times the derivative of g of x. The way I remember this is this chant. If you think of f of x as the high function, and g of x is a low function, you can say this is low D high minus high D low
6:03:27
minus f of x times the derivative of g of x. The way I remember this is this chant. If you think of f of x as the high function, and g of x is a low function, you can say this is low D high minus high D low over low low, over low low, we're low low means the low function squared. Let's start with a pretty simple example. So this derivative, taking us back to z here, we're low low means the low function squared. Let's start with a pretty simple example.
6:04:11
So this derivative, taking us back to z here, we put the low low on the bottom, and then we go low, we put the low low on the bottom, and then we go low, D high. D high. z squared is to z minus high, D low the derivative of Z's cubed plus one is three z squared plus zero, don't really need to write the zero there, I can simplify here a little bit to z to the fourth plus two z minus three z to the fourth,
6:04:41
z squared is to z minus high, D low the derivative of Z's cubed plus one is three z squared plus zero, don't really need to write the zero there, I can simplify here a little bit to z to the fourth plus two z minus three z to the fourth, over I'm over I'm not going to bother multiplying out this denominator, I think it looks simpler with it factored. So when I cancel things on the numerator, I'm getting to z minus z to the fourth over
6:05:14
z cubed plus one squared as the derivative of my quotient. So in this video, we saw the product rule and the quotient rule. I've written the rules here using the prime notation instead of the dy dx notation, but you should check that the formulas are really the same as before. For the proofs of these fabulously useful rules, you'll have to watch the next video. In this video, I'll prove the product rule and the quotient rule, along with one more
6:05:46
related rule, called the reciprocal rule. First, the proof of the product rule. To find the derivative of the product, f of x times g of x, I'm going to start as usual with the limit definition of derivative. So the limit as h goes to zero of f of x plus h g of x plus h minus f of x g of x, Oliver H. Now I'd like to make this expression here look more like the expression above it, which is where I'm heading for. And to do this, I'm
6:06:22
going to use a classic trick of adding zero to my expression and kind of a devious way. So I'm going to rewrite my expression, leaving the first term and the last term of the numerator as they are, but inserting two new terms that cancel. So I'm subtracting the term, f of x, g of x plus h, and then adding it back again. So they won't change the value of my expression. This is not as pointless as it seems, because now we can factor out the common
6:06:58
factor of g of x plus h from the first two terms, and the common factor of f of x from the next two terms. So I'm going to do that. And now I'll rewrite again, splitting up my sum into two pieces here. You can use just algebra fractions to see that this expression here and this expression here are the same. Notice that I'm taking the limit of this entire expression here. Now my limit rules allow me to rewrite this limit as four separate
6:07:40
limits, provided that these four limits do in fact exist, which that tell you show you in a moment that they do. So this first limit here is just equal to g of x, because G is a continuous function. G is continuous because by assumption it's differentiable, so it has to be continuous. This second limit here, you'll recognize as the definition of the derivative of f, so that limit exists at equals d dx of f of x. The third limit, well, f of
6:08:19
x has nothing to do with age. So that limit is just f of x. And finally, the fourth limit is the derivative of g. And we've done it. Well, modulus some minor rearrangement, you'll see that this expression here, is exactly the same as this expression here, just with the order of the terms switched around. Before we go on to prove the quotient rule, it'll be really handy to prove the reciprocal rule, which states that the derivative of reciprocal,
6:08:49
one over f of x is given by negative the derivative of f of x divided by f of x squared. So to prove this fact, let's start as usual, with the definition of derivative. The derivative of one over f of x is the limit as h goes to zero of one over f of not going to bother multiplying out this denominator, I think it looks simpler with it factored. So when I cancel things on the numerator, I'm getting to z minus z to the fourth over
6:09:30
z cubed plus one squared as the derivative of my quotient. So in this video, we saw the product rule and the quotient rule. I've written the rules here using the prime notation instead of the dy dx notation, but you should check that the formulas are really the same as before. For the proofs of these fabulously useful rules, you'll have to watch the next video. In this video, I'll prove the product rule and the quotient rule, along with one more
6:10:02
related rule, called the reciprocal rule. First, the proof of the product rule. To find the derivative of the product, f of x times g of x, I'm going to start as usual with the limit definition of derivative. So the limit as h goes to zero of f of x plus h g of x plus h minus f of x g of x, Oliver H. Now I'd like to make this expression here look more like the expression above it, which is where I'm heading for. And to do this, I'm
6:10:36
going to use a classic trick of adding zero to my expression and kind of a devious way. So I'm going to rewrite my expression, leaving the first term and the last term of the numerator as they are, but inserting two new terms that cancel. So I'm subtracting the term, f of x, g of x plus h, and then adding it back again. So they won't change the value of my expression. This is not as pointless as it seems, because now we can factor out the common
6:11:16
factor of g of x plus h from the first two terms, and the common factor of f of x from the next two terms. So I'm going to do that. And now I'll rewrite again, splitting up my sum into two pieces here. You can use just algebra fractions to see that this expression here and this expression here are the same. Notice that I'm taking the limit of this entire expression here. Now my limit rules allow me to rewrite this limit as four separate
6:11:47
limits, provided that these four limits do in fact exist, which that tell you show you in a moment that they do. So this first limit here is just equal to g of x, because G is a continuous function. G is continuous because by assumption it's differentiable, so it has to be continuous. This second limit here, you'll recognize as the definition of the derivative of f, so that limit exists at equals d dx of f of x. The third limit, well, f of
6:12:18
x has nothing to do with age. So that limit is just f of x. And finally, the fourth limit is the derivative of g. And we've done it. Well, modulus some minor rearrangement, you'll see that this expression here, is exactly the same as this expression here, just with the order of the terms switched around. Before we go on to prove the quotient rule, it'll be really handy to prove the reciprocal rule, which states that the derivative of reciprocal,
6:12:50
one over f of x is given by negative the derivative of f of x divided by f of x squared. So to prove this fact, let's start as usual, with the definition of derivative. The derivative of one over f of x is the limit as h goes to zero of one over f of x plus x plus h minus one over f of x over h. Now, these fractions here are just crying out to be combined by finding a common denominator that common denominators f of x plus h times f of x. So
6:13:32
let me do that. I've just multiplied the first fraction by f of x over f of x and the second fraction by f of x plus h over f of x plus h in order to rewrite with a common denominator. So that gives me f of x minus f of x plus h divided by f of x plus h times f of x. And instead of dividing this whole thing by H, I'll multiply it by one of our age, which gives me another factor of H in the denominator. Now, this expression here is looking a lot like the derivative of
6:14:09
h minus one over f of x over h. Now, these fractions here are just crying out to be combined by finding a common denominator that common denominators f of x plus h times f of x. So let me do that. I've just multiplied the first fraction by f of x over f of x and the second fraction by f of x plus h over f of x plus h in order to rewrite with a common denominator. So that gives me f of x minus f of x plus h divided by f of x plus h times f of x. And
6:14:43
instead of dividing this whole thing by H, I'll multiply it by one of our age, which gives me another factor of H in the denominator. Now, this expression here is looking a lot like the derivative of f. f. It's just in the reverse order. So let me factor out a negative sign in order to let me switch that order here. So this becomes f of x plus h minus f of x over h, and then I've got the times one over f of x plus h times f of x. Now I can split this limit up
6:15:22
first, I'll factor out the negative sign. And then I'll write this product as a product of two limits, which I can do provided the component limits exist. And I'll check that these limits do exist. Let's see the first limit here. The first limit here is just the derivative of It's just in the reverse order. So let me factor out a negative sign in order to let me switch that order here. So this becomes f of x plus h minus f of x over h, and then
6:15:52
I've got the times one over f of x plus h times f of x. Now I can split this limit up first, I'll factor out the negative sign. And then I'll write this product as a product of two limits, which I can do provided the component limits exist. And I'll check that these limits do exist. Let's see the first limit here. The first limit here is just the derivative of f. f. And the second limit here exists because f is continuous, f is continuous since is differentiable.
6:16:37
So by continuity, as H is going to zero, since x plus h is approaching x, f of x plus h is just approaching f of x. And I can rewrite this limit as one over f of x times f of x. And so this is in other words, negative the derivative of f of x divided by f of x squared. Now we've proved the reciprocal rule. Now we're in a great position to prove the quotient rule with very little effort. So instead of going back to the definition of derivative,
6:17:26
this time, I'm just gonna think of the quotient f of x over g of x as a product of f of x times the reciprocal And the second limit here exists because f is continuous, f is continuous since is differentiable. So by continuity, as H is going to zero, since x plus h is approaching x, f of x plus h is just approaching f of x. And I can rewrite this limit as one over f of x times f of x. And so this is in other words, negative the derivative of f of x divided by f of x squared.
6:18:21
Now we've proved the reciprocal rule. Now we're in a great position to prove the quotient rule with very little effort. So instead of going back to the definition of derivative, this time, I'm just gonna think of the quotient f of x over g of x as a product of f of x times the reciprocal of g of x. of g of x. And now, by the product rule, that's just the first function times the derivative of the second plus the derivative of the first times the second. And by the quotient rule,
6:19:11
the derivative of this reciprocal is negative derivative of g over g of x squared. And I still have this second term here, which I'm just going to write as derivative f of x divided by g of x. And now, by the product rule, that's just the first function times the derivative of the second plus the derivative of the first times the second. And by the quotient rule, the derivative of this reciprocal is negative derivative of g over g of x squared. And I
6:19:47
still have this second term here, which I'm just going to write as derivative f of x divided by g of x. Okay, Okay, so we're almost there. If we combine these two fractions, using a common denominator of g of x squared, we just have to multiply this second fraction by g of x over g of x to get that common denominator. Now we get negative f of x times the derivative of g of x plus the derivative of f of x times g of x divided by g of x squared. And hopefully,
6:20:28
this bottom expression is the same as this top expression. And, yes, after rearranging the terms it is. So that's the end of the proof of the quotient rule. So this video gave proofs of the product rule, the reciprocal rule, and then the quotient rule. This video is about two limits involving trig functions that turn out to be very useful. Namely, the limit as theta goes to zero of sine theta so we're almost there. If we combine these two fractions, using a common denominator
6:21:04
of g of x squared, we just have to multiply this second fraction by g of x over g of x to get that common denominator. Now we get negative f of x times the derivative of g of x plus the derivative of f of x times g of x divided by g of x squared. And hopefully, this bottom expression is the same as this top expression. And, yes, after rearranging the terms it is. So that's the end of the proof of the quotient rule. So this video
6:21:44
gave proofs of the product rule, the reciprocal rule, and then the quotient rule. This video is about two limits involving trig functions that turn out to be very useful. Namely, the limit as theta goes to zero of sine theta over theta. And the limit as theta goes to zero of cosine theta minus one over theta. These limits turn out to have really nice simple answers, as long as we keep theta in radians, over theta. And the limit as theta goes to zero of cosine theta minus one over theta.
6:22:25
These limits turn out to have really nice simple answers, as long as we keep theta in radians, not degrees. Let's consider the limit on the left first, that's the limit as theta goes to zero of sine theta over theta. Notice that we can't just evaluate this limit by plugging in zero for theta, because as theta goes to zero sine theta, and the numerator also goes to zero, and theta itself goes to zero, so we end up with a zero over zero indeterminate
6:22:54
form. We can however, build up some evidence of what this limit by might be by using a calculator and a table of values, or by looking at a graph. So here's the theta axis. And here's the y axis. And you can see that as theta goes to zero from either the right or the left, it's looking like the y value is going to one. The second limit here on the right, is also zero over zero and determinant form. Since as theta goes to zero, cosine
6:23:25
theta goes to one, so cosine theta minus one goes to zero. But again, looking at the graph, we have some evidence to suggest that as theta goes to zero, our expression is also going to not degrees. Let's consider the limit on the left first, that's the limit as theta goes to zero of sine theta over theta. Notice that we can't just evaluate this limit by plugging in zero for theta, because as theta goes to zero sine theta, and the numerator also goes
6:23:51
to zero, and theta itself goes to zero, so we end up with a zero over zero indeterminate form. We can however, build up some evidence of what this limit by might be by using a calculator and a table of values, or by looking at a graph. So here's the theta axis. And here's the y axis. And you can see that as theta goes to zero from either the right or the left, it's looking like the y value is going to one. The second limit here on the
6:24:36
right, is also zero over zero and determinant form. Since as theta goes to zero, cosine theta goes to one, so cosine theta minus one goes to zero. But again, looking at the graph, we have some evidence to suggest that as theta goes to zero, our expression is also going to zero. zero. these graphs provide strong evidence, but graphs can be misleading, and they're no substitute for a rigorous proof. So for a pretty cool geometric and algebraic proof of these facts,
6:25:15
please see the proof video for this section. The fact that the limit as theta goes to zero of sine theta over theta is one is really handy when you want to approximate sine theta. Because intuitively, this is saying that sine theta is approximately equal to theta itself when theta is near zero, because the ratio is approximately one. So if I want to approximate sine of this value of theta, without a calculator, I can use that fact. And say that the sine
6:25:51
of 0.01769 is going to be approximately equal to 0.01769. This is an important time to remind you that when we're doing these limits, we're assuming that theta is in radians. If it's not in radians, we won't get this nice limit of one here. So that's our approximation. And we can check it on a calculator, and I actually get an exact value of this number up to 10 decimal places. So as you can see, this is a really good approximation. We can
6:26:28
use this same limit fat, again, in the next example, to calculate this complicated limit as x goes to zero, the limit of tan of 7x over sine of forex. So when I see tangents and signs and expression, I'm always tempted to rewrite things. And just in terms of sine and cosine, so I'm going to do that first I'm going to rewrite tangent as as sine over cosine. That still divided by sine of forex, and now I'm going to flip and multiply to
6:27:01
get sine of 7x over cosine of 7x times one over sine of 4x. Now intuitively, if x is near zero, therefore 7x and 4x are also near zero, then sine of 7x is approximately equal to 7x. And sine of 4x is approximately equal to 4x. So intuitively, this limit should be pretty much the same thing as the limit as x goes to zero of 7x over cosine 7x times 4x. And canceling the access, this is just the same as seven fourths times the limit
6:27:42
of one of our cosine 7x. Since cosine of 7x is going to one, this should just be seven fourths. So this is the intuitive approach, let me also give you a more rigorous approach. So more rigorously, I'm going to rewrite this limit by multiplying by 7x over 7x. And by multiplying by 4x, over 4x, that hasn't changed my expression, I'm just multiplying by one and fancy forms. But this is really useful. Because if I regroup here, and write the sine
6:28:12
7x over the 7x, times the one over cosine 7x. Now I'm going to write the Forex over the sign for x. And I'm still left with a 7x from the top and a 4x. From the bottom here, I can cancel out those x's. And I can notice that this limit here, as x goes to 07, x is going to zero, so sine 7x over 7x is just going to be equal to one. And similarly as x goes to zero, for x is going to zero. So the limit of 4x ever signed for x is the
6:28:48
reciprocal of one, it's also one. And finally, this limit in the middle here, as x goes to 07, x is going to zero, so cosine of 7x is going to one, and everything in the world is going to one except the 7/4. So this limit is seven forth. these graphs provide strong evidence, but graphs can be misleading, and they're no substitute for a rigorous proof. So for a pretty cool geometric and algebraic proof of these facts,
6:29:19
please see the proof video for this section. The fact that the limit as theta goes to zero of sine theta over theta is one is really handy when you want to approximate sine theta. Because intuitively, this is saying that sine theta is approximately equal to theta itself when theta is near zero, because the ratio is approximately one. So if I want to approximate sine of this value of theta, without a calculator, I can use that fact. And say that the sine
6:30:00
of 0.01769 is going to be approximately equal to 0.01769. This is an important time to remind you that when we're doing these limits, we're assuming that theta is in radians. If it's not in radians, we won't get this nice limit of one here. So that's our approximation. And we can check it on a calculator, and I actually get an exact value of this number up to 10 decimal places. So as you can see, this is a really good approximation. We can
6:30:32
use this same limit fat, again, in the next example, to calculate this complicated limit as x goes to zero, the limit of tan of 7x over sine of forex. So when I see tangents and signs and expression, I'm always tempted to rewrite things. And just in terms of sine and cosine, so I'm going to do that first I'm going to rewrite tangent as as sine over cosine. That still divided by sine of forex, and now I'm going to flip and multiply to
6:31:13
get sine of 7x over cosine of 7x times one over sine of 4x. Now intuitively, if x is near zero, therefore 7x and 4x are also near zero, then sine of 7x is approximately equal to 7x. And sine of 4x is approximately equal to 4x. So intuitively, this limit should be pretty much the same thing as the limit as x goes to zero of 7x over cosine 7x times 4x. And canceling the access, this is just the same as seven fourths times the limit
6:31:43
of one of our cosine 7x. Since cosine of 7x is going to one, this should just be seven fourths. So this is the intuitive approach, let me also give you a more rigorous approach. So more rigorously, I'm going to rewrite this limit by multiplying by 7x over 7x. And by multiplying by 4x, over 4x, that hasn't changed my expression, I'm just multiplying by one and fancy forms. But this is really useful. Because if I regroup here, and write the sine
6:32:25
7x over the 7x, times the one over cosine 7x. Now I'm going to write the Forex over the sign for x. And I'm still left with a 7x from the top and a 4x. From the bottom here, I can cancel out those x's. And I can notice that this limit here, as x goes to 07, x is going to zero, so sine 7x over 7x is just going to be equal to one. And similarly as x goes to zero, for x is going to zero. So the limit of 4x ever signed for x is the
6:33:04
reciprocal of one, it's also one. And finally, this limit in the middle here, as x goes to 07, x is going to zero, so cosine of 7x is going to one, and everything in the world is going to one except the 7/4. So this limit is seven forth. In this video, we found that the limit as theta goes to zero of sine theta over theta is equal to one. And the limit as theta goes to zero of cosine theta minus one over theta
6:33:51
is equal to zero. There's a nice proof of these facts in a later video for this section. When you compose two functions, you apply the first function. And then you apply the second function to the output of the first function. For example, the first function might compute population size from time in years. In this video, we found that the limit as theta goes to zero of sine theta over theta is equal to one. And the limit as theta goes to zero of cosine theta minus one over theta
6:34:24
is equal to zero. There's a nice proof of these facts in a later video for this section. When you compose two functions, you apply the first function. And then you apply the second function to the output of the first function. For example, the first function might compute population size from time in years. So its input would be time in years, since a certain date, as output would be number of people in the population. The second function g, might compute health care costs as a function
6:34:55
of population size. So it will take population size as input, and its output will be healthcare costs. If you put these functions together, that is compose them, then you'll go all the way from time in years to healthcare costs. This is your composition, g composed with F. The composition of two functions, written g with a little circle, f of x is defined as follows. g composed with f of x is G evaluated on f of x, we can think of it schematically
6:35:19
and so diagram, f acts on a number x and produces a number f of x, then g takes that output f of x and produces a new number, g of f of x. Our composition of functions g composed with F is the function that goes all the way from X to g of f of x. Let's work out some examples where our functions are defined by tables of values. If we want to find g composed with F of four, by definition, this means g of f of four. To evaluate this expression,
6:35:56
we always work from the inside out. So we start with the x value of four, and we find f of four using the table of values for f of x. When x equals four, f of x is seven, so we can replace F of four with the number seven. Now we need to evaluate g of seven, seven becomes our new x value in our table of values for G, the x value of seven corresponds to the G of X value of 10. So g of seven is equal to 10. We found that g composed with
6:36:52
F of four is equal to 10. If instead we want to find f composed with g of four, well, we can rewrite that is f of g of four, and again, work from the inside out. Now we're trying to find g of four, so four is our x value. And we use our table of values for G to see that g of four is one. So we replaced by a four by one, and now we need to evaluate f of one. Using our table for F values, f of one is eight. Notice that when we've computed
6:37:25
g of f of four, we got a different answer there when we computed F of G, F four. And in general, g composed with F is not the same thing as f composed with g. So its input would be time in years, since a certain date, as output would be number of people in the population. The second function g, might compute health care costs as a function of population size. So it will take population size as input, and its output will be healthcare
6:37:58
costs. If you put these functions together, that is compose them, then you'll go all the way from time in years to healthcare costs. This is your composition, g composed with F. The composition of two functions, written g with a little circle, f of x is defined as follows. g composed with f of x is G evaluated on f of x, we can think of it schematically and so diagram, f acts on a number x and produces a number f of x, then g takes that output
6:38:34
f of x and produces a new number, g of f of x. Our composition of functions g composed with F is the function that goes all the way from X to g of f of x. Let's work out some examples where our functions are defined by tables of values. If we want to find g composed with F of four, by definition, this means g of f of four. To evaluate this expression, we always work from the inside out. So we start with the x value of four, and we find
6:39:07
f of four using the table of values for f of x. When x equals four, f of x is seven, so we can replace F of four with the number seven. Now we need to evaluate g of seven, seven becomes our new x value in our table of values for G, the x value of seven corresponds to the G of X value of 10. So g of seven is equal to 10. We found that g composed with F of four is equal to 10. If instead we want to find f composed with g of four, well, we
6:39:42
can rewrite that is f of g of four, and again, work from the inside out. Now we're trying to find g of four, so four is our x value. And we use our table of values for G to see that g of four is one. So we replaced by a four by one, and now we need to evaluate f of one. Using our table for F values, f of one is eight. Notice that when we've computed g of f of four, we got a different answer there when we computed F of G, F four. And
6:40:18
in general, g composed with F is not the same thing as f composed with g. Please Please pause the video and take a moment to compute the next two examples. We can replace f composed with F of two by the equivalent expression, f of f of two. Working from the inside out, we know that f of two is three, and f of three is six. If we want to find f composed with g of six, rewrite that as f of g of six isn't the table for g, g of six is eight. But F
6:40:59
of eight, eight is not on the table as an x value for the for the f function. And so there is no F of eight, this does not exist, we can say that six is not in the domain, for F composed with g. Even though it was in the domain of g, we couldn't follow all the way through and get a value for F composed with g of six. Next, let's turn our attention to the composition of functions that are given by equations.
6:41:25
pause the video and take a moment to compute the next two examples. We can replace f composed with F of two by the equivalent expression, f of f of two. Working from the inside out, we know that f of two is three, and f of three is six. If we want to find f composed with g of six, rewrite that as f of g of six isn't the table for g, g of six is eight. But F of eight, eight is not on the table as an x value for the for the f function. And so
6:41:58
there is no F of eight, this does not exist, we can say that six is not in the domain, for F composed with g. Even though it was in the domain of g, we couldn't follow all the way through and get a value for F composed with g of six. Next, let's turn our attention to the composition of functions that are given by equations. p of x is x squared plus x and q of x is negative 2x. We want to find q composed with P of one.
6:42:36
p of x is x squared plus x and q of x is negative 2x. We want to find q composed with P of one. As usual, I can rewrite this as Q of P of one and work from the inside out. P of one is one squared plus one, so that's two. So this is the same thing as Q of two. But queue of two is negative two times two or negative four. So this evaluates to negative four. In this next example, we want to find q composed with P of some arbitrary x, or rewrite it
6:43:16
as usual as Q of p of x and work from the inside out. Well, p of x, we know the formula for that, that's x squared plus x. So I can replace my P of x with that expression. Now, I'm stuck with evaluating q on x squared plus x. Well queue of anything is negative two times that thing. So q of x squared plus x is going to be negative two times the quantity x squared plus x, what I've done is I've substituted in the whole expression x squared plus x,
6:43:52
where I saw the X in this formula for q of x, it's important to use the parentheses here, so that we'll be multiplying negative two by the whole expression and not just by the first piece, I can simplify this a bit as negative 2x squared minus 2x. And that's my expression for Q composed with p of x. Notice that if I wanted to compute q composed with P of one, which I already did in the first problem, I could just use this expression
6:44:24
now, negative two times one squared minus two and I get negative four, just like I did before. Let's try another one. Let's try p composed with q of x. First I read right this P of q of x. Working from the inside out, I can replace q of x with negative 2x. So I need to compute P of negative 2x. Here's my formula for P. to compute P of this expression, I need to plug in this expression everywhere I see an x in the formula for P. So that means
6:45:02
negative 2x squared plus negative 2x. Again, being careful to use parentheses to make sure I plug in the entire expression in for x. let me simplify. This is 4x squared minus 2x. Notice that I got different expressions for Q of p of x. And for P of q of x. Once again, we see that q composed with P is not necessarily equal to P composed with Q. Please pause the video and try this last example yourself rewriting and working from the inside
6:45:44
out, we're going to replace p of x with its expression x squared plus x. And then we need to evaluate p on x squared As usual, I can rewrite this as Q of P of one and work from the inside out. P of one is one squared plus one, so that's two. So this is the same thing as Q of two. But queue of two is negative two times two or negative four. So this evaluates to negative four. In this next example, we want to find q composed with P of some arbitrary x, or rewrite it
6:46:30
as usual as Q of p of x and work from the inside out. Well, p of x, we know the formula for that, that's x squared plus x. So I can replace my P of x with that expression. Now, I'm stuck with evaluating q on x squared plus x. Well queue of anything is negative two times that thing. So q of x squared plus x is going to be negative two times the quantity x squared plus x, what I've done is I've substituted in the whole expression x squared plus x,
6:47:16
where I saw the X in this formula for q of x, it's important to use the parentheses here, so that we'll be multiplying negative two by the whole expression and not just by the first piece, I can simplify this a bit as negative 2x squared minus 2x. And that's my expression for Q composed with p of x. Notice that if I wanted to compute q composed with P of one, which I already did in the first problem, I could just use this expression
6:47:58
now, negative two times one squared minus two and I get negative four, just like I did before. Let's try another one. Let's try p composed with q of x. First I read right this P of q of x. Working from the inside out, I can replace q of x with negative 2x. So I need to compute P of negative 2x. Here's my formula for P. to compute P of this expression, I need to plug in this expression everywhere I see an x in the formula for P. So that means
6:48:38
negative 2x squared plus negative 2x. Again, being careful to use parentheses to make sure I plug in the entire expression in for x. let me simplify. This is 4x squared minus 2x. Notice that I got different expressions for Q of p of x. And for P of q of x. Once again, we see that q composed with P is not necessarily equal to P composed with Q. Please pause the video and try this last example yourself rewriting and working from the inside
6:49:09
out, we're going to replace p of x with its expression x squared plus x. And then we need to evaluate p on x squared plus x. plus x. That means we plug in That means we plug in x squared plus x, x squared plus x, everywhere we see an x in this formula, so that's x squared plus x quantity squared plus x squared plus x. Once again, I can simplify by distributing out, that gives me x to the fourth plus 2x cubed plus x squared plus x squared plus x, or x to the fourth plus 2x
6:49:58
cubed plus 2x squared plus x. In this last set of examples, we're asked to go backwards, we're given a formula for a function of h of x. But we're supposed to rewrite h of x as a composition of two functions, F and G. Let's think for a minute, which of these two functions gets applied first, f composed with g of x, let's see, that means f of g of x. And since we evaluate these expressions from the inside out, we must be applying g first,
6:50:30
and then F. In order to figure out what what f and g could be, I like to draw a box around some thing inside my expression for H, so I'm going to draw a box around x squared plus seven, then whatever's inside the box, that'll be my function, g of x, the first function that gets applied, whatever happens to the box, in this case, taking the square root sign, that becomes my outside function, my second function f. So here, we're gonna say
6:50:58
g of x is equal to x squared plus seven, and f of x is equal to the square root of x, let's just check and make sure that this works. So I need to check that when I take the composition, f composed with g, I need to get the same thing as my original h. So let's see, if I do f composed with g of x, well, by definition, that's f of g of x. everywhere we see an x in this formula, so that's x squared plus x quantity squared plus
6:51:43
x squared plus x. Once again, I can simplify by distributing out, that gives me x to the fourth plus 2x cubed plus x squared plus x squared plus x, or x to the fourth plus 2x cubed plus 2x squared plus x. In this last set of examples, we're asked to go backwards, we're given a formula for a function of h of x. But we're supposed to rewrite h of x as a composition of two functions, F and G. Let's think for a minute, which of these two
6:52:33
functions gets applied first, f composed with g of x, let's see, that means f of g of x. And since we evaluate these expressions from the inside out, we must be applying g first, and then F. In order to figure out what what f and g could be, I like to draw a box around some thing inside my expression for H, so I'm going to draw a box around x squared plus seven, then whatever's inside the box, that'll be my function, g of x, the first function
6:53:07
that gets applied, whatever happens to the box, in this case, taking the square root sign, that becomes my outside function, my second function f. So here, we're gonna say g of x is equal to x squared plus seven, and f of x is equal to the square root of x, let's just check and make sure that this works. So I need to check that when I take the composition, f composed with g, I need to get the same thing as my original h. So let's see, if I
6:53:57
do f composed with g of x, well, by definition, that's f of g of x. Working from the inside out, I can replace g of x with its formula x squared plus seven. So I need to evaluate f of x squared plus seven. That means I plug in x squared plus seven, into the formula for for F. So that becomes the square root of x squared plus seven, two, it works because it matches my original equation. So we found a correct answer
6:54:36
a correct way of breaking h down as a composition of two functions. But I do want to point out, this is not the only correct answer. I'll write down my formula for H of X again, and this time, I'll put the box in a different place, I'll just box the x squared. If I did that, then my inside function, my first function, g of x would be x squared. And my second function is what happens Working from the inside out, I can replace g of x with its formula x squared plus seven.
6:55:22
So I need to evaluate f of x squared plus seven. That means I plug in x squared plus seven, into the formula for for F. So that becomes the square root of x squared plus seven, two, it works because it matches my original equation. So we found a correct answer a correct way of breaking h down as a composition of two functions. But I do want to point out, this is not the only correct answer. I'll write down my formula for H of X again, and
6:55:59
this time, I'll put the box in a different place, I'll just box the x squared. If I did that, then my inside function, my first function, g of x would be x squared. And my second function is what happens to the box. to the box. So my f of x is what happens to the box, and the box gets added seven to it, and taking the square root. So in other words, f of x is going to be the square root of x plus seven. Again, I can check that this works. If I do f composed with g of x, that's f
6:56:52
So my f of x is what happens to the box, and the box gets added seven to it, and taking the square root. So in other words, f of x is going to be the square root of x plus seven. Again, I can check that this works. If I do f composed with g of x, that's f of g of x. of g of x. So now g of x is x squared, so I'm taking f of x squared. When I plug in x squared for x, I do in fact get the square root of x squared plus seven. So this is an alternative correct
6:57:47
solution. In this video, we learn to evaluate the composition of functions. by rewriting it and working from the inside out. We also learn to break apart a complicated function into a composition of two functions by boxing one piece of the function and letting the first function applied in the composition. Let that be the inside of the box and the second function applied in the composition be whatever happens to the box.
6:58:23
So now g of x is x squared, so I'm taking f of x squared. When I plug in x squared for x, I do in fact get the square root of x squared plus seven. So this is an alternative correct solution. In this video, we learn to evaluate the composition of functions. by rewriting it and working from the inside out. We also learn to break apart a complicated function into a composition of two functions by boxing one piece of the function and letting the
6:59:00
first function applied in the composition. Let that be the inside of the box and the second function applied in the composition be whatever happens to the box. This video is about solving rational equations. A rational equation like this one equation that has rational expressions and that, in other words, an equation that has some variables in the denominator. There are several different approaches for solving a rational equation,
6:59:25
but they all start by finding the least common denominator. In this example, the denominators are x plus three and x, we can think of one as just having a denominator of one. Since the denominators don't have any factors in common, I can find the least common denominator just by multiplying them together. My next step is going to be clearing the denominator. By this, I mean that I multiply both sides of my equation by this least common denominator,
6:59:51
x plus three times x, I multiply on the left side of the equation, and I multiply by the same thing on the right side of the equation. Since I'm doing the same thing to both sides of the equation, I don't change the the value of the equation. Multiplying the least common denominator on both sides of the equation is equivalent to multiplying it by all three terms in the equation, I can see this when I multiply out, I'll rewrite the left side
7:00:24
the same as before, pretty much. And then I'll distribute the right side to get x plus three times x times one plus x plus three times x times one over x. So I've actually multiplied the least common denominator by all three terms of my equation. Now I can have a blast canceling things. The x plus three cancels with the x plus three on the denominator. The here are nothing cancels out because there's no denominator, and here
7:00:55
are the x in the numerator cancels with the x in the denominator. So I can rewrite my expression as x squared equals x plus three times x times one plus x plus three. Now I'm going to simplify. So I'll leave the x squared alone on this side, I'll distribute out x squared plus 3x plus x plus three, hey, look, the x squared is cancel on both sides. And so I get zero equals 4x plus three, so 4x is negative three, and x is negative three
7:01:53
fourths. Finally, I'm going to plug in my answer to check. This is a good idea for any kind of equation. But it's especially important for a rational equation because occasionally for rational equations, you'll get what's called extraneous solution solutions that don't actually work in your original equation because they make the denominator zero. Now, in this example, I don't think we're going to get the extraneous equations, because negative
7:02:23
three fourths is not going to make any of these denominators zero, so it should work out fine when I plug in. If I plug in, I get this, I can simplify the denominator here, negative three fourths plus three, three is 12, for sets becomes nine fourths, and this is one or flip and multiply to get minus four thirds. So here, I can simplify my complex fraction, it ends up being negative three nights, and one minus four thirds is negative
7:02:55
1/3. So that all seems to check out. And so my final answer is x equals negative three fourths. This next example looks a little trickier. And it is, but the same approach will work. First off, find the least common denominator. So here, my denominators are c minus five, c plus one, and C squared minus four c minus five, I'm going to factor that as C minus five times c plus one. Now, my least common denominator needs to have just
7:03:25
enough factors to that each of these denominators divided into it. So I need the factor c minus five, I need the factor c plus one. And now I've already got all the factors I need for this denominator. So here is my least common denominator. Next step is to clear the denominators. So I do this by multiplying both sides of the equation by my least common denominator. In fact, I can just multiply each of the three terms by this least common denominator. I
7:03:59
went ahead and wrote my third denominator in factored form to make it easier to see what cancels. Now canceling time dies, this dies, and both of those factors die. This video is about solving rational equations. A rational equation like this one equation that has rational expressions and that, in other words, an equation that has some variables in the denominator. There are several different approaches for solving a rational equation,
7:04:31
but they all start by finding the least common denominator. In this example, the denominators are x plus three and x, we can think of one as just having a denominator of one. Since the denominators don't have any factors in common, I can find the least common denominator just by multiplying them together. My next step is going to be clearing the denominator. By this, I mean that I multiply both sides of my equation by this least common denominator,
7:04:56
x plus three times x, I multiply on the left side of the equation, and I multiply by the same thing on the right side of the equation. Since I'm doing the same thing to both sides of the equation, I don't change the the value of the equation. Multiplying the least common denominator on both sides of the equation is equivalent to multiplying it by all three terms in the equation, I can see this when I multiply out, I'll rewrite the left side
7:05:36
the same as before, pretty much. And then I'll distribute the right side to get x plus three times x times one plus x plus three times x times one over x. So I've actually multiplied the least common denominator by all three terms of my equation. Now I can have a blast canceling things. The x plus three cancels with the x plus three on the denominator. The here are nothing cancels out because there's no denominator, and here
7:06:19
are the x in the numerator cancels with the x in the denominator. So I can rewrite my expression as x squared equals x plus three times x times one plus x plus three. Now I'm going to simplify. So I'll leave the x squared alone on this side, I'll distribute out x squared plus 3x plus x plus three, hey, look, the x squared is cancel on both sides. And so I get zero equals 4x plus three, so 4x is negative three, and x is negative three
7:06:48
fourths. Finally, I'm going to plug in my answer to check. This is a good idea for any kind of equation. But it's especially important for a rational equation because occasionally for rational equations, you'll get what's called extraneous solution solutions that don't actually work in your original equation because they make the denominator zero. Now, in this example, I don't think we're going to get the extraneous equations, because negative
7:07:19
three fourths is not going to make any of these denominators zero, so it should work out fine when I plug in. If I plug in, I get this, I can simplify the denominator here, negative three fourths plus three, three is 12, for sets becomes nine fourths, and this is one or flip and multiply to get minus four thirds. So here, I can simplify my complex fraction, it ends up being negative three nights, and one minus four thirds is negative
7:07:37
1/3. So that all seems to check out. And so my final answer is x equals negative three fourths. This next example looks a little trickier. And it is, but the same approach will work. First off, find the least common denominator. So here, my denominators are c minus five, c plus one, and C squared minus four c minus five, I'm going to factor that as C minus five times c plus one. Now, my least common denominator needs to have just
7:08:06
enough factors to that each of these denominators divided into it. So I need the factor c minus five, I need the factor c plus one. And now I've already got all the factors I need for this denominator. So here is my least common denominator. Next step is to clear the denominators. So I do this by multiplying both sides of the equation by my least common denominator. In fact, I can just multiply each of the three terms by this least common denominator. I
7:08:34
went ahead and wrote my third denominator in factored form to make it easier to see what cancels. Now canceling time dies, this dies, and both of those factors die. cancel out the denominators, the whole point of multiplying by the least common denominator, you're multiplying by something that's big enough to kill every single denominator so you don't have to deal with denominators anymore. Now I'm going to simplify by multiplying out.
7:09:03
So I get, let's see, c plus one times four c, that's four c squared plus four c, now I get minus just c minus five, and then over here, I get three c squared plus three, I can rewrite the minus quantity c minus five as a minus c plus five. And now I can subtract the three c squared from both sides to get just a C squared over here, and the four c minus c, that becomes a three C. And finally, I can subtract the three from both sides to
7:09:39
get c squared plus three c plus two equals zero. got myself a quadratic equation that looks like a nice one that factors. So this factors to C plus one times c plus two equals zero. So either c plus one is zero, or C plus two is zero, so C equals negative one, or C equals negative two. Now let's see, we need to still check our answers. Without even going to the trouble of calculating anything, I can see that C equals negative one is not
7:10:22
going to work, because if I plug it in to this denominator here, I get a denominator of zero, which doesn't make sense. So C equals minus one is an extraneous solution, it doesn't actually satisfy my original equation. And so I can just cross it right out, C equals negative two. I can go if I go ahead, and that doesn't make any of my denominators zero. So if I haven't made any mistakes, it should satisfy my original equation, but, but I'll
7:10:49
just plug it in to be sure. And after some simplifying, I get a true statement. So my final answer is C equals negative two. In this video, we solved a couple of rational equations, using the method of finding the least common denominator, and then clearing the denominator, we cleared the denominator by multiplying both sides of the equation by the least common denominator or equivalently. multiplying each of the terms by that denominator.
7:11:21
There's another equivalent method that some people prefer, it still starts out the same, we find the least common denominator, but then we write all the fractions over that least common denominator. So in this example, we'd still use the least common denominator of x plus three times x. cancel out the denominators, the whole point of multiplying by the least common denominator, you're multiplying by something that's big enough to kill every single denominator so
7:11:49
you don't have to deal with denominators anymore. Now I'm going to simplify by multiplying out. So I get, let's see, c plus one times four c, that's four c squared plus four c, now I get minus just c minus five, and then over here, I get three c squared plus three, I can rewrite the minus quantity c minus five as a minus c plus five. And now I can subtract the three c squared from both sides to get just a C squared over here, and the four c
7:12:18
minus c, that becomes a three C. And finally, I can subtract the three from both sides to get c squared plus three c plus two equals zero. got myself a quadratic equation that looks like a nice one that factors. So this factors to C plus one times c plus two equals zero. So either c plus one is zero, or C plus two is zero, so C equals negative one, or C equals negative two. Now let's see, we need to still check our answers. Without even going
7:12:49
to the trouble of calculating anything, I can see that C equals negative one is not going to work, because if I plug it in to this denominator here, I get a denominator of zero, which doesn't make sense. So C equals minus one is an extraneous solution, it doesn't actually satisfy my original equation. And so I can just cross it right out, C equals negative two. I can go if I go ahead, and that doesn't make any of my denominators zero.
7:13:24
So if I haven't made any mistakes, it should satisfy my original equation, but, but I'll just plug it in to be sure. And after some simplifying, I get a true statement. So my final answer is C equals negative two. In this video, we solved a couple of rational equations, using the method of finding the least common denominator, and then clearing the denominator, we cleared the denominator by multiplying both sides of the equation
7:13:55
by the least common denominator or equivalently. multiplying each of the terms by that denominator. There's another equivalent method that some people prefer, it still starts out the same, we find the least common denominator, but then we write all the fractions over that least common denominator. So in this example, we'd still use the least common denominator of x plus three times x. But our next step would be to write each of these rational expressions over that common
7:14:27
denominator by multiplying the top and the bottom by the appropriate things. So one, in order to get the common denominator of x plus 3x, I need to multiply the top and the bottom by x plus three times x, whenever x, I need to multiply the top and the bottom just by x plus three since that's what's missing from the denominator x. Now, if I simplify a little bit, let's say this is x squared over that common denominator, and here I have
7:15:06
just x plus three times x over that denominator, and here I have x plus three over that common denominator. Now add together my fractions on the right side, so they have a common denominator. So this is x plus three times x plus x plus three. And now I have two fractions that have that are equal, that have the same denominator, therefore, their numerators have to be equal also. So the next step is to set the numerators equal. So I get x squared is x plus three
7:15:41
times x plus x plus three. And if you look back at the previous way, we solve this equation, you'll recognize this equation. And so from here on we just continue as before. When choosing between these two methods, I personally tend to prefer the clear the denominators method, because it's a little bit less writing, you don't have to get rid of those denominators earlier. You don't have to write them as many times, but some people find this one a little
7:16:11
bit easier to remember, a little easier to understand either of these methods is fine. One last caution. Don't forget at the end, to check your solutions and eliminate any extraneous solutions. These will be solutions that make the denominators of your original equation. Go to zero. This video gives the derivative of sine cosine and other trig functions. A graph of the function y equals sine x is given in blue here, we can estimate the shape
7:16:42
of the derivative of sine x by looking at the slopes of the tangent lines. Here, when x equals zero, the tangent line has a positive slope of approximately one. As x increases to pi over two, the slope of the tangent line is still positive, but decreases to zero. Next, the slope turns negative more and more negative reaching a negative value of negative one, before returning again to zero. Continuing like this, we see that the graph of the derivative
7:17:06
y equals sine prime of x looks like the graph of y equals cosine x below. Please pause the video and do a similar exercise for the graph of y equals cosine of x below, that is use the graph of y equals cosine x to estimate the shape of the graph of y equals cosine prime of x. Notice that when x equals zero, the slope of the tangent line here is zero, that slope turns negative, and then reaches zero again before turning positive. So the
7:17:39
graph of the derivative should look something like this. This new blue graph looks like the vertical reflection of the blue graph above suggesting that the derivative of cosine of x is equal to the negative of sine of x. So we have graphical evidence that the derivative of sine x is equal to cosine of x, and the derivative of cosine of x is equal to negative sine of x. For proofs of these facts, please see the separate proof video for this section.
7:18:09
Once we have the derivatives of sine and cosine, we have the power to compute the derivatives of a lot of other trig functions as well. But our next step would be to write each of these rational expressions over that common denominator by multiplying the top and the bottom by the appropriate things. So one, in order to get the common denominator of x plus 3x, I need to multiply the top and the bottom by x plus three times x, whenever x, I need to multiply the top and the bottom
7:18:45
just by x plus three since that's what's missing from the denominator x. Now, if I simplify a little bit, let's say this is x squared over that common denominator, and here I have just x plus three times x over that denominator, and here I have x plus three over that common denominator. Now add together my fractions on the right side, so they have a common denominator. So this is x plus three times x plus x plus three. And now I have two fractions that have
7:19:12
that are equal, that have the same denominator, therefore, their numerators have to be equal also. So the next step is to set the numerators equal. So I get x squared is x plus three times x plus x plus three. And if you look back at the previous way, we solve this equation, you'll recognize this equation. And so from here on we just continue as before. When choosing between these two methods, I personally tend to prefer the clear the denominators method,
7:19:45
because it's a little bit less writing, you don't have to get rid of those denominators earlier. You don't have to write them as many times, but some people find this one a little bit easier to remember, a little easier to understand either of these methods is fine. One last caution. Don't forget at the end, to check your solutions and eliminate any extraneous solutions. These will be solutions that make the denominators of your original
7:20:17
equation. Go to zero. This video gives the derivative of sine cosine and other trig functions. A graph of the function y equals sine x is given in blue here, we can estimate the shape of the derivative of sine x by looking at the slopes of the tangent lines. Here, when x equals zero, the tangent line has a positive slope of approximately one. As x increases to pi over two, the slope of the tangent line is still positive, but decreases to zero.
7:20:54
Next, the slope turns negative more and more negative reaching a negative value of negative one, before returning again to zero. Continuing like this, we see that the graph of the derivative y equals sine prime of x looks like the graph of y equals cosine x below. Please pause the video and do a similar exercise for the graph of y equals cosine of x below, that is use the graph of y equals cosine x to estimate the shape of the graph of y equals cosine
7:21:33
prime of x. Notice that when x equals zero, the slope of the tangent line here is zero, that slope turns negative, and then reaches zero again before turning positive. So the graph of the derivative should look something like this. This new blue graph looks like the vertical reflection of the blue graph above suggesting that the derivative of cosine of x is equal to the negative of sine of x. So we have graphical evidence that the derivative
7:22:00
of sine x is equal to cosine of x, and the derivative of cosine of x is equal to negative sine of x. For proofs of these facts, please see the separate proof video for this section. Once we have the derivatives of sine and cosine, we have the power to compute the derivatives of a lot of other trig functions as well. And notice that a nice way to remember which of these answers have negative signs in them is that the derivatives of the trig functions that start with a co
7:22:38
And notice that a nice way to remember which of these answers have negative signs in them is that the derivatives of the trig functions that start with a co always have a negative, and the root of the trig functions that don't have the ko are positive. Now let's use these formulas in an example. g of x is a complicated expression involving several trig functions as well as a constant m, and I have a couple choices
7:23:03
of how to proceed. I could try to rewrite all my trig functions in terms of sine and cosine and simplify, or I could attack the derivative directly using the quotient rule. I'm going to use the direct approach In this case, but sometimes you'll find that rewriting will make things easier. So using the quotient rule on the denominator, I get the original denominator squared. On the numerator, I get low D high to compute the derivative of x
7:23:33
cosine x, I need the product rule. So I get x times the derivative of cosine, which is negative sine x, plus the derivative of x, which is just one times cosine of x. Now I have to do a minus Hi, x cosine of x dillow. The derivative of M is just zero because M is a constant, plus the derivative of cotangent which is negative cosecant squared of x. So I found the derivative, I'm going to go ahead and simplify a little bit by multiplying out
7:24:05
then rewriting everything in terms of sine and cosine, and then multiplying the numerator and denominator by sine squared of x, we have a somewhat simplified expression for the derivative, you should memorize the derivatives of the trig functions will prove that the first two formulas are correct in a separate proof video. In this video, I'll give proofs for the two special trig limit. And I'll also prove that the derivative of sine is cosine. And the
7:24:34
derivative of cosine is minus sign. To prove that the limit of sine that over theta is one as theta goes to zero, I'm going to start with a picture. In this picture, I have a unit circle a circle of radius one, and I have two right triangles, a green triangle and a smaller red triangle, both with angle theta. Now I'm going to argue in terms of areas, if I want to compute the area of this sector that I've shaded in blue here, in other
7:25:08
words, that pie shaped piece, I can first compute the area of the circle, which is pi times one squared for the radius. But since the sector has angle theta, and the full circle has angle two pi, I need to multiply that area of the circle by the ratio theta over two pi to represent the fraction of the area of the circle that's included in this sector. So in other words, the area of the sector is just going to be theta over two, where
7:25:49
theta is given in the radians. Now if I want to compute the area of the little red triangle, I can do one half times the base times the height. Now the base is going to be equal to cosine theta, because I have a circle of radius one angle theta here, and the height is going to be sine theta. Finally, the area of the green triangle is also one half times the base times the height. But now the base is a full one unit, and the height is given
7:26:38
by tangent theta, since opposite, which is the height here over adjacent, which is one has to equal tangent theta. Now if I put all those areas together, I know that the area of the red triangle, alright is cosine theta sine theta over two has to be less than or equal to the area of the blue sector, theta over two, which is less than or equal to the area of the big green triangle, which is tan theta over two. Now I'm going to multiply
7:27:22
through this inequality by two and rewrite things in terms of sine and cosine to get cosine theta sine theta is less than or equal to theta is less than or equal to sine theta over cosine theta. Now I'm going to divide through my inequalities by sine theta, which won't change the inequalities as long as theta is greater than zero, so that sine theta is positive. always have a negative, and the root of the trig functions that don't have the ko are
7:27:53
positive. Now let's use these formulas in an example. g of x is a complicated expression involving several trig functions as well as a constant m, and I have a couple choices of how to proceed. I could try to rewrite all my trig functions in terms of sine and cosine and simplify, or I could attack the derivative directly using the quotient rule. I'm going to use the direct approach In this case, but sometimes you'll find that rewriting
7:28:12
will make things easier. So using the quotient rule on the denominator, I get the original denominator squared. On the numerator, I get low D high to compute the derivative of x cosine x, I need the product rule. So I get x times the derivative of cosine, which is negative sine x, plus the derivative of x, which is just one times cosine of x. Now I have to do a minus Hi, x cosine of x dillow. The derivative of M is just zero because M
7:28:44
is a constant, plus the derivative of cotangent which is negative cosecant squared of x. So I found the derivative, I'm going to go ahead and simplify a little bit by multiplying out then rewriting everything in terms of sine and cosine, and then multiplying the numerator and denominator by sine squared of x, we have a somewhat simplified expression for the derivative, you should memorize the derivatives of the trig functions will prove that the first two
7:29:10
formulas are correct in a separate proof video. In this video, I'll give proofs for the two special trig limit. And I'll also prove that the derivative of sine is cosine. And the derivative of cosine is minus sign. To prove that the limit of sine that over theta is one as theta goes to zero, I'm going to start with a picture. In this picture, I have a unit circle a circle of radius one, and I have two right triangles, a green triangle
7:29:57
and a smaller red triangle, both with angle theta. Now I'm going to argue in terms of areas, if I want to compute the area of this sector that I've shaded in blue here, in other words, that pie shaped piece, I can first compute the area of the circle, which is pi times one squared for the radius. But since the sector has angle theta, and the full circle has angle two pi, I need to multiply that area of the circle by the ratio theta over
7:30:26
two pi to represent the fraction of the area of the circle that's included in this sector. So in other words, the area of the sector is just going to be theta over two, where theta is given in the radians. Now if I want to compute the area of the little red triangle, I can do one half times the base times the height. Now the base is going to be equal to cosine theta, because I have a circle of radius one angle theta here, and the height
7:30:59
is going to be sine theta. Finally, the area of the green triangle is also one half times the base times the height. But now the base is a full one unit, and the height is given by tangent theta, since opposite, which is the height here over adjacent, which is one has to equal tangent theta. Now if I put all those areas together, I know that the area of the red triangle, alright is cosine theta sine theta over two has to be less than or
7:31:42
equal to the area of the blue sector, theta over two, which is less than or equal to the area of the big green triangle, which is tan theta over two. Now I'm going to multiply through this inequality by two and rewrite things in terms of sine and cosine to get cosine theta sine theta is less than or equal to theta is less than or equal to sine theta over cosine theta. Now I'm going to divide through my inequalities by sine theta, which
7:32:22
won't change the inequalities as long as theta is greater than zero, so that sine theta is positive. And I get cosine theta is less than or equal to theta over sine theta is less than or equal to one over cosine theta. Now this middle expression is the reciprocal of the expression I want to take the limit of. So I'm going to go ahead and take limits. And since the limits of the two expressions on the outside, both exist, and equal one by the sandwich
7:33:19
theorem, the limit of the expression on the inside has to exist an equal one as well. Now I've cheated a little bit here. And I've really just taken the limit from the right because I've assumed that theta is greater than zero. But you can check that if you say that less than zero, so that sign that as negative, the limit from the left will also equal one, the inequalities will flip around first, but you'll still get it use the sandwich
7:34:04
theorem to get a limit of one. And that's a cool geometric proof of this useful limit from calculus. To show that the limit of cosine theta minus one over theta is zero, we can actually rewrite this expression and reuse the limit that we just computed. So let me write down my limit. And I'm going to multiply this expression by cosine theta plus one on the numerator and the denominator. So I haven't changed the expression, I'm just multiply that by one. Now, if I multiply my numerator out,
7:34:35
And I get cosine theta is less than or equal to theta over sine theta is less than or equal to one over cosine theta. Now this middle expression is the reciprocal of the expression I want to take the limit of. So I'm going to go ahead and take limits. And since the limits of the two expressions on the outside, both exist, and equal one by the sandwich theorem, the limit of the expression on the inside has to exist an equal one as well.
7:35:06
Now I've cheated a little bit here. And I've really just taken the limit from the right because I've assumed that theta is greater than zero. But you can check that if you say that less than zero, so that sign that as negative, the limit from the left will also equal one, the inequalities will flip around first, but you'll still get it use the sandwich theorem to get a limit of one. And that's a cool geometric proof of this useful limit
7:35:35
from calculus. To show that the limit of cosine theta minus one over theta is zero, we can actually rewrite this expression and reuse the limit that we just computed. So let me write down my limit. And I'm going to multiply this expression by cosine theta plus one on the numerator and the denominator. So I haven't changed the expression, I'm just multiply that by one. Now, if I multiply my numerator out, I get cosine squared theta minus one. And from the trig identity, sine squared theta
7:36:12
plus cosine squared theta equals one, I know that cosine squared theta minus one has to equal minus sine squared. So I can rewrite my limit as the limit of minus sine squared theta over theta cosine theta plus one. And now I can regroup to write my sine theta over theta, and my other copy of sine theta over cosine theta plus one, the limit of the first expression is going to be negative one, because of the limit we just proved. And the limit
7:36:48
of the second expression is just zero over one plus one, or zero, and therefore, my entire limit is just going to be negative one times zero, or zero, which is exactly what we want it to prove. Now we can use these two limits that we've just proved to calculate the derivatives of sine and cosine, using the limit definition of derivative and prove the results that were stated previously. According to the limit definition of derivative, the derivative of
7:37:32
sine x is the limit as h goes to zero of sine of x plus h minus sine of x divided by H. As usual, this is a zero for zero indeterminate form limits. So I'm going to need to rewrite things to evaluate it. And I'm going to rewrite using the angle sum formula for sine, the sine of x plus h is equal to sine x cosine H, cosine plus cosine x sine H. Now if I rearrange things, and factor out a sine x from the first term, I can break up my limit into pieces
7:38:20
and compute every piece. So this is sine x times zero plus cosine x times one. And so my final answer is cosine x as we wanted the proof that the derivative of cosine is minus sine is very similar. So please stop the video and try it for yourself before proceeding. Using the limit definition of derivative, we have that the derivative of cosine of x is the limit as h goes to zero of cosine of x plus h minus cosine of x over h, we can
7:38:55
rewrite the cosine of x plus h using the angle sum formula as the cosine of x times the cosine of H minus the sine of x times the sine of H. And then we still have the minus cosine of x over h. As before, we're going to regroup things and factoring out the cosine x from the first part, the same familiar limits just put together in different ways. So here, cosine of x as h goes to zero is just cosine of x. This limit we know is zero. sine of x is just
7:39:31
saying sine of x, and sine of h over h is going to one which means that our final answer is going to be negative sine of x times one or just negative sine of x, which is exactly what we wanted. That's all for the proofs of these four useful calculus facts. rectilinear motion or linear motion means the motion of an object along a straight line. For example, a particle moving left and right, or a ball going up and down. In this video, we'll see
7:40:17
what the derivative and the second derivative tell us about the motion of an object constrained to move along a straight line. In this example, A particle is moving up and down along a straight line. And its position is given by this equation where the positive positions mean that the particle is above its Baseline Position, whatever I'm calling position zero, and negative positions mean the particle is below this Baseline Position. I'm asked to find s prime of T and S double
7:40:54
prime of t. So by deriving I get four t cubed minus 16 t squared plus 12 T for the first derivative, and 12 t squared minus 32 t plus 12. For the second derivative, I get cosine squared theta minus one. And from the trig identity, sine squared theta plus cosine squared theta equals one, I know that cosine squared theta minus one has to equal minus sine squared. So I can rewrite my limit as the limit of minus sine squared
7:41:27
theta over theta cosine theta plus one. And now I can regroup to write my sine theta over theta, and my other copy of sine theta over cosine theta plus one, the limit of the first expression is going to be negative one, because of the limit we just proved. And the limit of the second expression is just zero over one plus one, or zero, and therefore, my entire limit is just going to be negative one times zero, or zero, which is exactly what we want
7:42:02
it to prove. Now we can use these two limits that we've just proved to calculate the derivatives of sine and cosine, using the limit definition of derivative and prove the results that were stated previously. According to the limit definition of derivative, the derivative of sine x is the limit as h goes to zero of sine of x plus h minus sine of x divided by H. As usual, this is a zero for zero indeterminate form limits. So I'm going to need to rewrite
7:42:37
things to evaluate it. And I'm going to rewrite using the angle sum formula for sine, the sine of x plus h is equal to sine x cosine H, cosine plus cosine x sine H. Now if I rearrange things, and factor out a sine x from the first term, I can break up my limit into pieces and compute every piece. So this is sine x times zero plus cosine x times one. And so my final answer is cosine x as we wanted the proof that the derivative of cosine is minus
7:43:16
sine is very similar. So please stop the video and try it for yourself before proceeding. Using the limit definition of derivative, we have that the derivative of cosine of x is the limit as h goes to zero of cosine of x plus h minus cosine of x over h, we can rewrite the cosine of x plus h using the angle sum formula as the cosine of x times the cosine of H minus the sine of x times the sine of H. And then we still have the minus cosine
7:43:50
of x over h. As before, we're going to regroup things and factoring out the cosine x from the first part, the same familiar limits just put together in different ways. So here, cosine of x as h goes to zero is just cosine of x. This limit we know is zero. sine of x is just saying sine of x, and sine of h over h is going to one which means that our final answer is going to be negative sine of x times one or just negative sine of x, which is exactly
7:44:40
what we wanted. That's all for the proofs of these four useful calculus facts. rectilinear motion or linear motion means the motion of an object along a straight line. For example, a particle moving left and right, or a ball going up and down. In this video, we'll see what the derivative and the second derivative tell us about the motion of an object constrained to move along a straight line. In this example, A particle is moving up and down along a straight
7:45:10
line. And its position is given by this equation where the positive positions mean that the particle is above its Baseline Position, whatever I'm calling position zero, and negative positions mean the particle is below this Baseline Position. I'm asked to find s prime of T and S double prime of t. So by deriving I get four t cubed minus 16 t squared plus 12 T for the first derivative, and 12 t squared minus 32 t plus 12. For the second derivative,
7:45:42
S prime of t, which can also be written, D STD represents the instantaneous rate of change of S of t, the position over time, well, the change in position over time is just the velocity. And this can also be written as v of t, s double prime of t, the second derivative of s with respect to t, can also be thought of as the derivative of the velocity function. So that represents the rate of change of velocity over time, how fast the velocity is increasing
7:46:27
or decreasing. And that is called acceleration. And it can be written as a lefty. Like position, velocity and acceleration can be both positive and negative. A positive velocity means the position is increasing. So the particle is moving up, while a negative velocity means the position is decreasing, so the particle is moving down. Of course, a velocity of zero means the particles at rest, at least for that instant. from physics, we know that force
7:47:05
equals mass times acceleration. So if the acceleration is positive, then that means the force is in the positive direction, it's like the particle is being pulled up. If on the other hand, the acceleration is negative than the force is in the negative direction, and it's like the particle is being pulled down. an acceleration of zero means there's no force on the particle at that instant, and the velocity continues as is. Let's use
7:47:33
these ideas about velocity acceleration. And the following table of values to describe the particles motion at time equals 1.5 seconds. At time 1.5 seconds, the position of the particle is positive, so that means the particle is above its Baseline Position of zero. its velocity is negative, so that means that its position is decreasing. In other words, the particle is moving down. Its acceleration is negative acceleration is the derivative of velocity.
7:48:07
So a negative acceleration means the velocity is decreasing. Well, a negative velocity that's decreasing is getting more and more negative. So in fact, the particle is moving down faster and faster. This can be a little bit confusing, because even though the velocity is decreasing, it's getting more and more negative, the speed, which is the absolute value of velocity is increasing. We can also see what the particle is doing at 1.5 seconds by looking at this
7:48:36
graph, where the time is drawn on the x axis and position s of t is on the y axis. From the graph, we can see that when t is zero, S of t is also zero. So the particle starts at its Baseline Position of zero. At time 1.5 seconds, the particle is above this starting position, but moving downwards. And since the slope of this graph is getting steeper and steeper, we can conclude that the speed of the particle is increasing. Same thing
7:49:13
was we concluded from the table of values. Now let's do the same analysis when time is 2.5 seconds. s of 2.5 seconds is negative. So the particle is below its starting position. Velocity s prime of t is also negative. So the particle is still going down. But now the acceleration as double prime of t is positive. That means that the velocity Today is increasing, well, a negative velocity that's increasing is getting less negative closer to zero. So
7:49:41
the particle must be slowing down. And in fact, the speed is decreasing. Again, the graph agrees with this reasoning, at 2.5 seconds, our position is way down here, our graph is decreasing, so the particles moving down, and the slope seems to be leveling off. So the particle speed is decreasing. Even though it's velocity, S prime of t, which can also be written, D STD represents the instantaneous rate of change
7:50:10
of S of t, the position over time, well, the change in position over time is just the velocity. And this can also be written as v of t, s double prime of t, the second derivative of s with respect to t, can also be thought of as the derivative of the velocity function. So that represents the rate of change of velocity over time, how fast the velocity is increasing or decreasing. And that is called acceleration. And it can be written as a lefty. Like position,
7:51:12
velocity and acceleration can be both positive and negative. A positive velocity means the position is increasing. So the particle is moving up, while a negative velocity means the position is decreasing, so the particle is moving down. Of course, a velocity of zero means the particles at rest, at least for that instant. from physics, we know that force equals mass times acceleration. So if the acceleration is positive, then that means
7:51:46
the force is in the positive direction, it's like the particle is being pulled up. If on the other hand, the acceleration is negative than the force is in the negative direction, and it's like the particle is being pulled down. an acceleration of zero means there's no force on the particle at that instant, and the velocity continues as is. Let's use these ideas about velocity acceleration. And the following table of values to describe
7:52:12
the particles motion at time equals 1.5 seconds. At time 1.5 seconds, the position of the particle is positive, so that means the particle is above its Baseline Position of zero. its velocity is negative, so that means that its position is decreasing. In other words, the particle is moving down. Its acceleration is negative acceleration is the derivative of velocity. So a negative acceleration means the velocity is decreasing. Well, a negative velocity that's
7:52:44
decreasing is getting more and more negative. So in fact, the particle is moving down faster and faster. This can be a little bit confusing, because even though the velocity is decreasing, it's getting more and more negative, the speed, which is the absolute value of velocity is increasing. We can also see what the particle is doing at 1.5 seconds by looking at this graph, where the time is drawn on the x axis and position s of t is on the y axis. From
7:53:24
the graph, we can see that when t is zero, S of t is also zero. So the particle starts at its Baseline Position of zero. At time 1.5 seconds, the particle is above this starting position, but moving downwards. And since the slope of this graph is getting steeper and steeper, we can conclude that the speed of the particle is increasing. Same thing was we concluded from the table of values. Now let's do the same analysis when time is
7:54:01
2.5 seconds. s of 2.5 seconds is negative. So the particle is below its starting position. Velocity s prime of t is also negative. So the particle is still going down. But now the acceleration as double prime of t is positive. That means that the velocity Today is increasing, well, a negative velocity that's increasing is getting less negative closer to zero. So the particle must be slowing down. And in fact, the speed is decreasing. Again, the
7:54:33
graph agrees with this reasoning, at 2.5 seconds, our position is way down here, our graph is decreasing, so the particles moving down, and the slope seems to be leveling off. So the particle speed is decreasing. Even though it's velocity, which you can think of as speed with direction is increasing, simply because it's a negative velocity that's getting less negative. Notice that in the first example, when velocity and
7:55:08
acceleration are both in the same direction, that is, they're both negative the particle was speeding up. And the second example, were velocity and acceleration when the opposite directions one positive one negative, the particle is slowing down. which you can think of as speed with direction is increasing, simply because it's a negative velocity that's getting less negative. Notice that in the first example, when velocity and
7:55:36
acceleration are both in the same direction, that is, they're both negative the particle was speeding up. And the second example, were velocity and acceleration when the opposite directions one positive one negative, the particle is slowing down. This is true in general, when velocity acceleration had the same sign, that as they're both positive or both negative, then the particle is speeding up. This is true in general, when velocity acceleration had the same sign, that as they're both positive
7:56:18
or both negative, then the particle is speeding up. And when velocity acceleration have opposite signs, then the particle is slowing down. One way to think about this is in terms of force, forces in the same direction as acceleration. So if velocity acceleration have the same sign, that means force is the same direction as the particles already going, so it's making the particle speed up. But if velocity and acceleration have opposite signs, then the force is going against the way that particles
7:57:00
moving, so it's causing it to slow down. Let's continue the same example with some more questions, it'll be helpful to write down the velocity and acceleration functions that we calculated earlier. I've also graphed position, velocity and acceleration here at the right. And before you go on, it's a fun exercise to figure out which one is which, without even looking at the equations just based on the shapes of the graphs. And where they're increasing where
7:57:29
they're decreasing where they're positive and where they're negative. Velocity is the derivative of position. So velocity needs to be positive, where position is increasing. The only pairs of functions that have this property are the blue one, that's positive, when the red ones increasing, and the green function, which is positive, when the blue one is increasing. Now acceleration, which is the derivative of velocity also needs to
7:57:58
be positive, when velocity is increasing. So the only way to correctly label the functions with both of these relationships is to make the red one be position, the blue one be velocity, and the green one be acceleration. This agrees with the equations that And when velocity acceleration have opposite signs, then the particle is slowing down. One way to think about this is in terms of force, forces in the same direction as acceleration.
7:58:28
So if velocity acceleration have the same sign, that means force is the same direction as the particles already going, so it's making the particle speed up. But if velocity and acceleration have opposite signs, then the force is going against the way that particles moving, so it's causing it to slow down. Let's continue the same example with some more questions, it'll be helpful to write down the velocity and acceleration functions that we calculated
7:59:02
earlier. I've also graphed position, velocity and acceleration here at the right. And before you go on, it's a fun exercise to figure out which one is which, without even looking at the equations just based on the shapes of the graphs. And where they're increasing where they're decreasing where they're positive and where they're negative. Velocity is the derivative of position. So velocity needs to be positive, where position is increasing.
7:59:42
The only pairs of functions that have this property are the blue one, that's positive, when the red ones increasing, and the green function, which is positive, when the blue one is increasing. Now acceleration, which is the derivative of velocity also needs to be positive, when velocity is increasing. So the only way to correctly label the functions with both of these relationships is to make the red one be position, the blue one be velocity,
8:00:38
and the green one be acceleration. This agrees with the equations that we have over here. we have over here. The first question asks, When is the particle at rest, the particle is temporarily at rest when the velocity is zero. In other words, S prime of t is zero. So plugging in the equation for S prime of t, we can factor out a four T, and factor some more. To conclude the T has to be 01, or three. The first question asks, When is the particle at rest, the particle is temporarily at rest
8:01:18
when the velocity is zero. In other words, S prime of t is zero. So plugging in the equation for S prime of t, we can factor out a four T, and factor some more. To conclude the T has to be 01, or three. This conclusion agrees with our graph of V of t, which has x intercepts at 01 and three, and also agrees with our graph of position s of t. Since the particle stops for a moment, it changed direction, when t equals 01. And three, the particle is moving up when velocity
8:02:06
is positive, and moving down when velocity is negative. Since we know from the previous question, the velocity equals zero, when t equals 01. And three, we can look in between those values to figure out whether the velocity is positive or negative, just by plugging in values. So for example, when t is negative one, if I plug into the negative one to the equation for velocity, I get a negative number. So velocity must be negative when t is less
8:02:39
than zero, between zero and one. If I plug in, for example, t equals one half, I get a value of S prime of t or V of t of 2.5, which is a positive number. If I plug in a value of t in between one and three, say t equals two, I get a value of v of t of negative eight, which is a negative number. And finally, if I plug in a value of t greater than three, so For, I get a positive answer for V of t. So from the sign chart, I see that V of t
8:03:14
is negative when t is between negative infinity and zero, and in between one and three, and V of t is positive when t is between This conclusion agrees with our graph of V of t, which has x intercepts at 01 and three, and also agrees with our graph of position s of t. Since the particle stops for a moment, it changed direction, when t equals 01. And three, the particle is moving up when velocity is positive, and moving down when velocity is negative. Since we know from the previous
8:03:52
question, the velocity equals zero, when t equals 01. And three, we can look in between those values to figure out whether the velocity is positive or negative, just by plugging in values. So for example, when t is negative one, if I plug into the negative one to the equation for velocity, I get a negative number. So velocity must be negative when t is less than zero, between zero and one. If I plug in, for example, t equals one half, I get
8:04:32
a value of S prime of t or V of t of 2.5, which is a positive number. If I plug in a value of t in between one and three, say t equals two, I get a value of v of t of negative eight, which is a negative number. And finally, if I plug in a value of t greater than three, so For, I get a positive answer for V of t. So from the sign chart, I see that V of t is negative when t is between negative infinity and zero, and in between one and three, and V of t is positive when t is between zero and
8:05:19
zero and one, and between three and infinity. Of course, I could have reached the same conclusion just by looking at the graph of velocity and where it's above and below the x axis, or even by looking at the graph of position and seeing where it's increasing and where it's decreasing. To answer the next question, the particle will be speeding up when V of t and a of t are both positive or both negative. And the particle will be slowing down when V of t
8:05:57
and a of t have opposite signs. So let's make a similar sign chart to figure out where a of t is positive and negative. First, it'll be helpful to find out where a if t is zero. So if I set zero equal to my S double prime, that's 12 t squared minus 32 t plus 12, I could factor out one, and between three and infinity. Of course, I could have reached the same conclusion just by looking at the graph of velocity and where it's above and below the x axis, or even by
8:06:33
looking at the graph of position and seeing where it's increasing and where it's decreasing. To answer the next question, the particle will be speeding up when V of t and a of t are both positive or both negative. And the particle will be slowing down when V of t and a of t have opposite signs. So let's make a similar sign chart to figure out where a of t is positive and negative. First, it'll be helpful to find out where a if t is zero.
8:07:25
So if I set zero equal to my S double prime, that's 12 t squared minus 32 t plus 12, I could factor out a four a four and then use the quadratic equation to find the solution. Since this equation doesn't factor easily, this simplifies to four thirds plus or minus the square root of seven over three, which is approximately 0.45, and 2.22. Now I can build a similar sign chart for acceleration, mark the places where acceleration is zero. and plug in values of t, say t equals zero
8:08:10
is the equation for acceleration, I get a positive answer here. When I plug in, say t equals one, I get a negative answer here. And when I plug in, say, t equals three, I get another positive answer here. Now, if I put this together with my velocity chart, which changed sign at 01, and three, and went from negative to positive, to negative to positive, I can try to figure out where velocity acceleration both have the same sign. It might
8:08:44
be helpful actually to shade in where acceleration is positive. And separately shade in where velocity is positive. And then look for the places where both are shaded. So between zero and 4.5, and greater than three. And then I can also look where both are unshaded that looks like in between one and 2.22. So that's where they're both negative. And then I'll know that V of t and a of t have opposite signs everywhere else. A slightly better answer,
8:09:24
we'll use exact values of four thirds plus or minus squared of seven thirds instead of these decimal approximations. So let me write that down. So here is where the particle speeding up. And here, it's where it's slowing down, we can check our work by looking at the graph of position, the particle speeding up with the position graph is getting steeper and steeper, that's the red graph is getting steeper and steeper here, here. And here, just like
8:09:58
we found algebraically. As our final example, let's look at net change in position and distance traveled between one and four seconds for the same particle. At time, one second position is five thirds, or about 1.67 millimeters at time for its position is given by 32 thirds, or about 10.67 millimeters, all I'm doing is plugging one and four into this equation. So the net change in position is just the difference of these two numbers. As a four
8:10:37
minus as of one, which is nine millimeters. At first glance, it might seem like the total distance traveled between one and four seconds should also be nine millimeters, but actually, it's a little more complicated. Because the particle switches direction during that time period, it doesn't go straight from its position at one second to its position at four seconds. Remember what the graph Position looked like the particle switches direction at one second
8:11:08
and at three seconds. So to find the total distance, we need the distance of travels from one second to three seconds, plus the distance that travels from three seconds to four seconds. Another way of thinking about this is that we need the absolute value of s three minus s one, plus the absolute value of s four minus s three, we need these absolute value signs because this difference in position will be negative instead of positive when
8:11:59
the particles moving down. Plugging in the t values into our equation, we get negative 27 minus five thirds, plus the absolute value of 32 thirds minus negative 27, which is a total of 199 thirds, or 66.3 repeating millimeters, quite a bit more than the nine millimeter difference in position. This video gave an in depth analysis of a particle moving up and down along a straight and then use the quadratic equation to find the solution. Since this equation doesn't
8:12:27
factor easily, this simplifies to four thirds plus or minus the square root of seven over three, which is approximately 0.45, and 2.22. Now I can build a similar sign chart for acceleration, mark the places where acceleration is zero. and plug in values of t, say t equals zero is the equation for acceleration, I get a positive answer here. When I plug in, say t equals one, I get a negative answer here. And when I plug in, say, t equals three, I
8:12:56
get another positive answer here. Now, if I put this together with my velocity chart, which changed sign at 01, and three, and went from negative to positive, to negative to positive, I can try to figure out where velocity acceleration both have the same sign. It might be helpful actually to shade in where acceleration is positive. And separately shade in where velocity is positive. And then look for the places where both are shaded. So between zero
8:13:22
and 4.5, and greater than three. And then I can also look where both are unshaded that looks like in between one and 2.22. So that's where they're both negative. And then I'll know that V of t and a of t have opposite signs everywhere else. A slightly better answer, we'll use exact values of four thirds plus or minus squared of seven thirds instead of these decimal approximations. So let me write that down. So here is where the particle speeding
8:13:55
up. And here, it's where it's slowing down, we can check our work by looking at the graph of position, the particle speeding up with the position graph is getting steeper and steeper, that's the red graph is getting steeper and steeper here, here. And here, just like we found algebraically. As our final example, let's look at net change in position and distance traveled between one and four seconds for the same particle. At time, one second position
8:14:16
is five thirds, or about 1.67 millimeters at time for its position is given by 32 thirds, or about 10.67 millimeters, all I'm doing is plugging one and four into this equation. So the net change in position is just the difference of these two numbers. As a four minus as of one, which is nine millimeters. At first glance, it might seem like the total distance traveled between one and four seconds should also be nine millimeters, but actually,
8:14:47
it's a little more complicated. Because the particle switches direction during that time period, it doesn't go straight from its position at one second to its position at four seconds. Remember what the graph Position looked like the particle switches direction at one second and at three seconds. So to find the total distance, we need the distance of travels from one second to three seconds, plus the distance that travels from three seconds to
8:15:11
four seconds. Another way of thinking about this is that we need the absolute value of s three minus s one, plus the absolute value of s four minus s three, we need these absolute value signs because this difference in position will be negative instead of positive when the particles moving down. Plugging in the t values into our equation, we get negative 27 minus five thirds, plus the absolute value of 32 thirds minus negative 27, which is a
8:15:48
total of 199 thirds, or 66.3 repeating millimeters, quite a bit more than the nine millimeter difference in position. This video gave an in depth analysis of a particle moving up and down along a straight line. line. A similar analysis could be done for a particle moving left and right along a straight line, where a positive position means the particles on the right side, and a negative position means the particles on the left side of his Baseline Position. Of course, the same analysis
8:16:15
can be done for other objects, not just particles. A typical application is to a ball being thrown straight up and then falling down again. This video will give an economic application of the derivative to a cost function. Suppose that the total cost of producing x tie dyed t shirts is C of x. A similar analysis could be done for a particle moving left and right along a straight line, where a positive position means the particles on the right side, and a negative position
8:16:56
means the particles on the left side of his Baseline Position. Of course, the same analysis can be done for other objects, not just particles. A typical application is to a ball being thrown straight up and then falling down again. This video will give an economic application of the derivative to a cost function. Suppose that the total cost of producing x tie dyed t shirts is C of x. I'm going to sketch a few graphs, and you try to decide which graph is the most reasonable
8:17:35
representation for C of x. Pause the video for a moment to think about it. All of these candidate graphs that I've drawn have a nonzero y intercept, that's meant to reflect the idea that there's some fixed startup cost and buying equipment before you can even get started. Now, I would like to suggest that C of x should be an increasing function of x, because it's going to cost more money to make more t shirts, you need more supplies and labor. So this
8:18:01
function is out. Now it's somewhat reasonable, I think that C of x might be a linear function of x like it is here, if you've got the same cost per t shirt, whether you make 10 t shirts, or 1000 t shirts, the slope in that case would represent the cost per t shirt. And the linear function would mean that cost per t shirt is constant, no matter how many t shirts you're making. But in reality, it's probably going to be cheaper to make 1000 t shirts than it
8:18:31
is to make just a few t shirts. And therefore the cost per t shirt, sure slope should be going down as x increases. So this function right here is the one whose slope is going down for larger access. And so I would say that this is the most reasonable representation for C of x as a function of x. In other words, C of x should be an increasing function, but C prime of x should be decreasing. C of 204 minus C of 200 represents the additional cost
8:19:03
for making 204 t shirts instead of 200. In formula, you might think of that as the cost of making the last four t shirts. The ratio C of 200 for a minus C of 200 over four is the average rate of change of C of x. The units are units of cost, which is probably dollars per t shirt. And formula you might think of this as the additional cost per t shirt of making the last four t shirts. C prime of 200 is the instantaneous rate of
8:19:39
change of C of x. c of x is known as the cost function. And C prime of x is called the marginal cost, which is the rate at which cost is increasing per additional t shirt made. It might seem a little bit weird to take the derivative of C of x since x can really only take on integer values. But we can always approximate C of x with a function whose domain is all real numbers. To make this a little more specific, let's use a cost function of C of x equals
8:20:19
500 plus 300 times the square root of x. In this example, x is supposed to be the number of iPads that are produced, and C of x is the cost of producing them in dollars. Then C of 401 minus C of 400. given by 500 plus 300, times the square root of 401 minus 500 plus 300 times the square root of 400. This simplifies to $7.50, rounded to the nearest cent. This means that it costs an additional $7.50 to go from producing 400 iPads to 401 iPad. In this fictitious example,
8:20:49
I'm going to sketch a few graphs, and you try to decide which graph is the most reasonable representation for C of x. Pause the video for a moment to think about it. All of these candidate graphs that I've drawn have a nonzero y intercept, that's meant to reflect the idea that there's some fixed startup cost and buying equipment before you can even get started. Now, I would like to suggest that C of x should be an increasing function of x, because it's
8:21:17
going to cost more money to make more t shirts, you need more supplies and labor. So this function is out. Now it's somewhat reasonable, I think that C of x might be a linear function of x like it is here, if you've got the same cost per t shirt, whether you make 10 t shirts, or 1000 t shirts, the slope in that case would represent the cost per t shirt. And the linear function would mean that cost per t shirt is constant, no matter how many t shirts you're
8:21:44
making. But in reality, it's probably going to be cheaper to make 1000 t shirts than it is to make just a few t shirts. And therefore the cost per t shirt, sure slope should be going down as x increases. So this function right here is the one whose slope is going down for larger access. And so I would say that this is the most reasonable representation for C of x as a function of x. In other words, C of x should be an increasing function, but
8:22:20
C prime of x should be decreasing. C of 204 minus C of 200 represents the additional cost for making 204 t shirts instead of 200. In formula, you might think of that as the cost of making the last four t shirts. The ratio C of 200 for a minus C of 200 over four is the average rate of change of C of x. The units are units of cost, which is probably dollars per t shirt. And formula you might think of this as the additional cost per t
8:23:00
shirt of making the last four t shirts. C prime of 200 is the instantaneous rate of change of C of x. c of x is known as the cost function. And C prime of x is called the marginal cost, which is the rate at which cost is increasing per additional t shirt made. It might seem a little bit weird to take the derivative of C of x since x can really only take on integer values. But we can always approximate C of x with a function whose domain is all
8:23:28
real numbers. To make this a little more specific, let's use a cost function of C of x equals 500 plus 300 times the square root of x. In this example, x is supposed to be the number of iPads that are produced, and C of x is the cost of producing them in dollars. Then C of 401 minus C of 400. given by 500 plus 300, times the square root of 401 minus 500 plus 300 times the square root of 400. This simplifies to $7.50, rounded to the nearest
8:23:58
cent. This means that it costs an additional $7.50 to go from producing 400 iPads to 401 iPad. In this fictitious example, if I want to compute C prime of 400, instead, I can see that C prime of x is equal to 300 times one half x to the minus one half. So C prime of 400 is going to be 300 times one half times 400 to the negative one half, which simplifies to 300 over two times the square root of 400, which is also 7.5, or $7.50. Per iPad. Up to rounding to the nearest cent,
8:24:35
these two answers are equal. And it makes sense that C prime of 400 should equal approximately this difference. Since C prime of 400, the derivative is approximately equal to the average rate of change going from 400 to 401, which is just this difference, divided by one. Once again, C prime of 400 is called the marginal cost, and represents the rate at which the cost function is increasing with each additional item. This video gave an example of the cost
8:25:07
function, and it's derivative, which is known as marginal cost. This video introduces logarithms. logarithms are a way of writing exponents. The expression log base a of B equals c means that a to the C equals b. In other words, log base a of B is the exponent that you raise a to to get be. The number A is called the base of the logarithm. It's also called the base when we write it in this exponential form. Some students find it helpful to remember
8:25:50
this relationship, log base a of B equals c means a to the C equals b, by drawing arrows, if I want to compute C prime of 400, instead, I can see that C prime of x is equal to 300 times one half x to the minus one half. So C prime of 400 is going to be 300 times one half times 400 to the negative one half, which simplifies to 300 over two times the square root of 400, which is also 7.5, or $7.50. Per iPad. Up to rounding to the nearest cent,
8:26:44
these two answers are equal. And it makes sense that C prime of 400 should equal approximately this difference. Since C prime of 400, the derivative is approximately equal to the average rate of change going from 400 to 401, which is just this difference, divided by one. Once again, C prime of 400 is called the marginal cost, and represents the rate at which the cost function is increasing with each additional item. This video gave an example of the cost
8:27:41
function, and it's derivative, which is known as marginal cost. This video introduces logarithms. logarithms are a way of writing exponents. The expression log base a of B equals c means that a to the C equals b. In other words, log base a of B is the exponent that you raise a to to get be. The number A is called the base of the logarithm. It's also called the base when we write it in this exponential form. Some students find it helpful to remember
8:28:36
this relationship, log base a of B equals c means a to the C equals b, by drawing arrows, a to the C equals b. a to the C equals b. Other students like to think of it in terms of asking a question, log base a of fee, asks, What power do you raise a two in order to get b? Let's look at some examples. log base two of eight is three, because two to the three equals eight. In general, log base two of y is asking you the question, What power do you have to raise to two to get y? So for
8:29:22
example, log base two of 16 is four, because it's asking you the question to two what power equals 16? And the answer is four. Please pause the video and try some of these other examples. log base two of two is asking, What power do you raise to two to get to? And the answer is one. Two to the one equals two. log base two of one half is asking two to what power gives you one half? Well, to get one half, you need to raise two to a negative
8:30:01
power. So that would be two to the negative one. So the answer is negative one. log base two of 1/8 means what power do we raise to two in order to get 1/8. Since one eight is one over two cubed, we have to raise two to the negative three power to get one over two cubed. So our exponent is negative three, and that's our answer to our log expression. Finally, log base two of one is asking to to what power equals one, or anything raised
8:30:30
to the zero power gives us one, so this log expression evaluates to zero. Notice that we can get positive negative and zero answers for our logarithm expressions. Please pause the video and figure out what these logs evaluate to to work out log base 10 of another Again, notice that a million is 10 to the sixth power. Now we're asking the question, What power do we raise tend to to get a million? So that is what power do we raise 10? to to get 10
8:30:57
to the six? Well, of course, the answer is going to be six. Similarly, since point O one is 10 to the minus three, this log expressions, the same thing as asking, what's the log base 10 of 10 to the minus three? Well, what power do you have to raise 10? to to get 10 to the minus three? Of course, the answer is negative three. Log base 10 of zero is asking, What power do we raise 10 to to get zero. If you think about it, there's no way to raise 10
8:31:42
to an exponent get zero. Raising 10 to a positive exponent gets us really big positive numbers. Raising 10 to a negative exponent is like one over 10 to a power that's giving us tiny fractions, but they're still positive numbers, we're never going to get zero. Even if we raised 10 to the zero power, we'll just get one. So there's no way to get zero and the log base 10 of zero does not exist. If you try it on your calculator using the log base
8:32:16
10 button, you'll get an error message. Same thing happens when we do log base 10 of negative 100. We're asking 10 to what power equals negative 100. And there's no exponent that will work. And more generally, it's possible to take the log of numbers that are greater than zero, but not for numbers that are less than or equal to zero. In other words, the domain of the function log base a of x, no matter what base you're using, for a, the
8:32:43
domain is going to be all positive numbers. A few notes on notation. When you see ln of x, that's called natural log, and it means the log base e of x, where he is that famous number that's about 2.718. When you see log of x with no base at all, by convention, that means log base 10 of x, and it's called the common log. Most scientific calculators have buttons for natural log, and for common log. Let's practice rewriting expressions with
8:33:16
logs in them. log base three of one nine is negative two can be rewritten as the expression three to the negative two equals 1/9. Log of 13 is shorthand for log base 10 of 13. So that can be rewritten as 10 to the 1.11394 equals 13. Finally, in this last expression, ln means natural log, or log base e, so I can rewrite this equation as log base e of whenever E equals negative one. Well, that means the same thing as e to the negative
8:33:58
one equals one over e, which is true. Now let's go the opposite direction. We'll start with exponential equations and rewrite them as logs. Remember that log base a of B equals c means the same thing Other students like to think of it in terms of asking a question, log base a of fee, asks, What power do you raise a two in order to get b? Let's look at some examples. log base two of eight is three, because two to the three equals eight. In general, log base two
8:34:39
of y is asking you the question, What power do you have to raise to two to get y? So for example, log base two of 16 is four, because it's asking you the question to two what power equals 16? And the answer is four. Please pause the video and try some of these other examples. log base two of two is asking, What power do you raise to two to get to? And the answer is one. Two to the one equals two. log base two of one half is asking two to
8:35:24
what power gives you one half? Well, to get one half, you need to raise two to a negative power. So that would be two to the negative one. So the answer is negative one. log base two of 1/8 means what power do we raise to two in order to get 1/8. Since one eight is one over two cubed, we have to raise two to the negative three power to get one over two cubed. So our exponent is negative three, and that's our answer to our log expression.
8:36:05
Finally, log base two of one is asking to to what power equals one, or anything raised to the zero power gives us one, so this log expression evaluates to zero. Notice that we can get positive negative and zero answers for our logarithm expressions. Please pause the video and figure out what these logs evaluate to to work out log base 10 of another Again, notice that a million is 10 to the sixth power. Now we're asking the question, What power
8:36:31
do we raise tend to to get a million? So that is what power do we raise 10? to to get 10 to the six? Well, of course, the answer is going to be six. Similarly, since point O one is 10 to the minus three, this log expressions, the same thing as asking, what's the log base 10 of 10 to the minus three? Well, what power do you have to raise 10? to to get 10 to the minus three? Of course, the answer is negative three. Log base 10 of zero is asking, What
8:37:16
power do we raise 10 to to get zero. If you think about it, there's no way to raise 10 to an exponent get zero. Raising 10 to a positive exponent gets us really big positive numbers. Raising 10 to a negative exponent is like one over 10 to a power that's giving us tiny fractions, but they're still positive numbers, we're never going to get zero. Even if we raised 10 to the zero power, we'll just get one. So there's no way to get zero and the
8:37:57
log base 10 of zero does not exist. If you try it on your calculator using the log base 10 button, you'll get an error message. Same thing happens when we do log base 10 of negative 100. We're asking 10 to what power equals negative 100. And there's no exponent that will work. And more generally, it's possible to take the log of numbers that are greater than zero, but not for numbers that are less than or equal to zero. In other words, the
8:38:35
domain of the function log base a of x, no matter what base you're using, for a, the domain is going to be all positive numbers. A few notes on notation. When you see ln of x, that's called natural log, and it means the log base e of x, where he is that famous number that's about 2.718. When you see log of x with no base at all, by convention, that means log base 10 of x, and it's called the common log. Most scientific calculators have
8:39:17
buttons for natural log, and for common log. Let's practice rewriting expressions with logs in them. log base three of one nine is negative two can be rewritten as the expression three to the negative two equals 1/9. Log of 13 is shorthand for log base 10 of 13. So that can be rewritten as 10 to the 1.11394 equals 13. Finally, in this last expression, ln means natural log, or log base e, so I can rewrite this equation as log base e of
8:39:56
whenever E equals negative one. Well, that means the same thing as e to the negative one equals one over e, which is true. Now let's go the opposite direction. We'll start with exponential equations and rewrite them as logs. Remember that log base a of B equals c means the same thing as a to the C equals b, as a to the C equals b, the base stays the same in both expressions. So for this example, the base of three in
8:40:32
the exponential equation, that's going to be the same as the base in our log. Now I just have to figure out what's in the argument of the log. And what goes on the other side of the equal sign. Remember that the answer to a log is an exponent. So the thing that goes in this box should be my exponent for my exponential equation. In other words, you and I'll put the 9.78 as the argument of my log. This works because log base three of
8:41:23
9.78 equals u means the same thing as three to the U equals 9.78, which is just what we started with. In the second example, the base of my exponential equation is E. So the base of my log is going to be the answer to my log is an exponent. In this case, the exponent 3x plus seven. And the other expression, the four minus y becomes my argument of my log. Let me check, log base e of four minus y equals 3x plus seven means e to 3x plus seven equals
8:42:06
four minus Y, which is just what I started with. I can also rewrite log base e as natural log. This video introduced the idea of logs. And the fact that log base a of B equal c means the same thing as a to the C equals b. So log base a of B is asking you the question, What power exponent Do you raise a to in order to get b. In this video, we'll work out the graph, so some log functions and also talk about their domains. For this first example,
8:42:50
let's graph a log function by hand by plotting some points. The function we're working with is y equals log base two of x, I'll make a chart of x and y values. Since we're working this out by hand, I want to pick x values for which it's easy to compute log base two of x. So I'll start out with the x value of one. Because log base two of one is zero, log base anything of one is 02 is another x value that's easy to compute log base two
8:43:22
of two, that's asking, What power do I raise to two to get to one? And the answer is one. Power other powers of two are easy to work with. So for example, log base two of four that saying what power do I raise to to to get four, so the answer is two. Similarly, log base two of eight is three, and log base two of 16 is four. Let me also work with some fractional values for X. If x is one half, then log base two of one half that saying
8:44:03
what power do I raise to two to get one half? Well, that needs a power of negative one. It's also easy to compute by hand, the log base two of 1/4 and 1/8. log base two of 1/4 is negative two, since two to the negative two is 1/4. And similarly, log base two of 1/8 is negative three. I'll put some tick marks on my x and y axes. Please pause the video and take a moment to plot these points. Let's see I have the point, one, zero, that's
8:44:54
here to one that's here, for two, that is here. And then eight, three, which is the base stays the same in both expressions. So for this example, the base of three in the exponential equation, that's going to be the same as the base in our log. Now I just have to figure out what's in the argument of the log. And what goes on the other side of the equal sign. Remember that the answer to a log is an exponent. So the thing that
8:45:38
goes in this box should be my exponent for my exponential equation. In other words, you and I'll put the 9.78 as the argument of my log. This works because log base three of 9.78 equals u means the same thing as three to the U equals 9.78, which is just what we started with. In the second example, the base of my exponential equation is E. So the base of my log is going to be the answer to my log is an exponent. In this case, the exponent
8:46:17
3x plus seven. And the other expression, the four minus y becomes my argument of my log. Let me check, log base e of four minus y equals 3x plus seven means e to 3x plus seven equals four minus Y, which is just what I started with. I can also rewrite log base e as natural log. This video introduced the idea of logs. And the fact that log base a of B equal c means the same thing as a to the C equals b. So log base a of B is asking you the question,
8:46:51
What power exponent Do you raise a to in order to get b. In this video, we'll work out the graph, so some log functions and also talk about their domains. For this first example, let's graph a log function by hand by plotting some points. The function we're working with is y equals log base two of x, I'll make a chart of x and y values. Since we're working this out by hand, I want to pick x values for which it's easy to compute log base two
8:47:21
of x. So I'll start out with the x value of one. Because log base two of one is zero, log base anything of one is 02 is another x value that's easy to compute log base two of two, that's asking, What power do I raise to two to get to one? And the answer is one. Power other powers of two are easy to work with. So for example, log base two of four that saying what power do I raise to to to get four, so the answer is two. Similarly,
8:47:54
log base two of eight is three, and log base two of 16 is four. Let me also work with some fractional values for X. If x is one half, then log base two of one half that saying what power do I raise to two to get one half? Well, that needs a power of negative one. It's also easy to compute by hand, the log base two of 1/4 and 1/8. log base two of 1/4 is negative two, since two to the negative two is 1/4. And similarly, log base two of
8:48:15
1/8 is negative three. I'll put some tick marks on my x and y axes. Please pause the video and take a moment to plot these points. Let's see I have the point, one, zero, that's here to one that's here, for two, that is here. And then eight, three, which is here. here. And the fractional x values, one half goes with negative one, and 1/4 with negative two 1/8 with negative three. And the fractional x values, one half goes with negative one, and 1/4 with negative two 1/8 with negative three.
8:48:50
And if I connect the dots, I get a graph that looks something like this. If I had smaller and smaller fractions, I would keep getting more and more negative answers when I took log base two of them, so my graph is getting more and more negative, my y values are getting more and more negative as x is getting close to zero. Now I didn't draw any parts of the graph over here with negative X values, I didn't put any negative X values on my chart.
8:49:13
That omission is no accident. Because if you try to take the log base two or base anything of a negative number, like say negative four or something, there's no answer. This doesn't exist because there's no power that you can raise to two to get a negative number. So there are no points on the graph for negative X values. And similarly, there are no points on the graph where x is zero, because you can't take log base two of zero, there's no
8:49:21
power you can raise to to to get zero. I want to observe some key features of this graph. First of all, the domain is x values greater than zero. In interval notation, I can write that as a round bracket because I don't want to include zero to infinity, the range is going to be the y values, while they go all the way down into the far reaches of the negative numbers. And the graph gradually increases y value is getting bigger and bigger. So the
8:49:29
range is actually all real numbers are an interval notation negative infinity to infinity. Finally, I want to point out that this graph has a vertical asymptote at the y axis, that is at the line x equals zero. I'll draw that on my graph with a dotted line. A vertical asymptote is a line that our functions graph gets closer and closer to so this is The graph of y equals log base two of x. But if I wanted to graph say, y equals log base 10 of x, it
8:49:35
would look very similar, it would still have a domain of X values greater than zero, a range of all real numbers and a vertical asymptote at the y axis, it would still go through the point one zero, but it would go through the point 10. One instead, because log base 10 of 10 is one, it would look pretty much the same, just a lot flatter over here. But even though it doesn't look like it with the way I've drawn it, it still gradually goes up
8:49:40
to n towards infinity. In fact, the graph of y equals log base neaa of X for a bigger than one looks pretty much the same, and has the same three properties. Now that we know what the basic log graph looks like, we can plot at least rough graphs of other log functions without plotting points. Here we have the graph of natural log of X plus five. And again, I'm just going to draw a rough graph. If I did want to do a more accurate graph, I probably
8:49:45
would want to plot some points. But I know that roughly a log graph, if it was just like y equals ln of x, that would look something like this, and it would go through the point one zero, And if I connect the dots, I get a graph that looks something like this. If I had smaller and smaller fractions, I would keep getting more and more negative answers when I took log base two of them, so my graph is getting more and more negative, my y values are getting
8:50:00
more and more negative as x is getting close to zero. Now I didn't draw any parts of the graph over here with negative X values, I didn't put any negative X values on my chart. That omission is no accident. Because if you try to take the log base two or base anything of a negative number, like say negative four or something, there's no answer. This doesn't exist because there's no power that you can raise to two to get a negative number. So
8:50:07
there are no points on the graph for negative X values. And similarly, there are no points on the graph where x is zero, because you can't take log base two of zero, there's no power you can raise to to to get zero. I want to observe some key features of this graph. First of all, the domain is x values greater than zero. In interval notation, I can write that as a round bracket because I don't want to include zero to infinity, the range is
8:50:19
going to be the y values, while they go all the way down into the far reaches of the negative numbers. And the graph gradually increases y value is getting bigger and bigger. So the range is actually all real numbers are an interval notation negative infinity to infinity. Finally, I want to point out that this graph has a vertical asymptote at the y axis, that is at the line x equals zero. I'll draw that on my graph with a dotted line. A vertical
8:50:24
asymptote is a line that our functions graph gets closer and closer to so this is The graph of y equals log base two of x. But if I wanted to graph say, y equals log base 10 of x, it would look very similar, it would still have a domain of X values greater than zero, a range of all real numbers and a vertical asymptote at the y axis, it would still go through the point one zero, but it would go through the point 10. One instead, because log base 10
8:50:29
of 10 is one, it would look pretty much the same, just a lot flatter over here. But even though it doesn't look like it with the way I've drawn it, it still gradually goes up to n towards infinity. In fact, the graph of y equals log base neaa of X for a bigger than one looks pretty much the same, and has the same three properties. Now that we know what the basic log graph looks like, we can plot at least rough graphs of other log functions
8:50:34
without plotting points. Here we have the graph of natural log of X plus five. And again, I'm just going to draw a rough graph. If I did want to do a more accurate graph, I probably would want to plot some points. But I know that roughly a log graph, if it was just like y equals ln of x, that would look something like this, and it would go through the point one zero, with a vertical asymptote along the y axis. Now if I want to graph ln of x plus five,
8:50:40
that just shifts our graph by five units, it'll still have the same vertical asymptote, since the vertical line shifted up by five units is still a vertical line. But instead of going through one zero, it'll go through the point, one, five. So I'll draw a rough sketch here. Let's compare our starting function y equals ln x, and the transformed version y equals ln x plus five in terms of the domain, the range and the vertical asymptote. Our
8:50:45
original function y equals ln x had a domain of zero to infinity. Since adding five on the outside affects the y values, and the domain is the x values, this transformation doesn't change the domain. So the domain is still from zero to infinity. Now the range of our original y equals ln x was from negative infinity to infinity. Shifting up by five does affect the y values, and the range is talking about the y values. But since the
8:50:50
original range was all real numbers, if you add five to all set of all real numbers, you still get the set of all real numbers. So in this case, the range doesn't change either. And finally, we already saw that the original vertical asymptote of the y axis x equals zero, when we shift that up by five units, it's still the vertical line x equals zero. In this next example, we're starting with a log base 10 function. And since the plus
8:50:55
two is on the inside, that means we shift that graph left by two. So I'll draw our basic log function. Here's our basic log function. So I'll think of that as y equals log of x going through the point one, zero, here's its vertical asymptote. Now I need to shift everything left by two. So my vertical asymptote shifts left, and now it's at the line x equals negative two, instead of at x equals zero, and my graph, let's see my point, one zero
8:51:00
gets shifted to, let's see negative one zero, since I'm subtracting two from the axis, and here's a rough sketch of the resulting graph. Let's compare the features of the two graphs drawn here. We're talking about domains, the original kind of domain of from zero to infinity. But now I've shifted that left. So I subtracted two from all my exercises. And here's my new domain, which I can also verify just by looking at the picture. My range was originally from
8:51:05
negative infinity to infinity. Well, shifting left only affects the x value, so it doesn't even affect the range. So my range is still negative infinity to infinity. My vertical asymptote was originally at x equals zero. And since I subtract two from all my x values, that shifts that to x equals negative two. In this last problem, I'm not going to worry about drawing this graph. I'll just use algebra to compute its domain. So let's think about
8:51:10
What's the issue, when you're taking the logs of things? Well, you can't take the log of a negative number or zero. So whatever is inside the argument of the log function, whatever is being fed into log had better be greater than zero. So I'll write that down, we need to minus 3x, to be greater than zero. Now it's a matter of solving an inequality. Two has got to be greater than 3x. So two thirds is greater than x. In other words, x has to
8:51:15
be less than two thirds. So our domain is all the x values from negative infinity to two thirds, not including two thirds, it's a good idea to memorize the basic shape of the graph of a log function. It looks something like this, go through the point one zero, and has a vertical asymptote on the y axis. Also, if you remember that you can't take the log of a negative number, or zero, then that helps you quickly compute domains for
8:51:20
log functions. Whatever's inside the log function, you set that greater than zero, and solve. This video is about combining logs and exponents. Please pause the video and take a moment to use your calculator to evaluate the following four expressions. with a vertical asymptote along the y axis. Now if I want to graph ln of x plus five, that just shifts our graph by five units, it'll still have the same vertical asymptote,
8:51:25
since the vertical line shifted up by five units is still a vertical line. But instead of going through one zero, it'll go through the point, one, five. So I'll draw a rough sketch here. Let's compare our starting function y equals ln x, and the transformed version y equals ln x plus five in terms of the domain, the range and the vertical asymptote. Our original function y equals ln x had a domain of zero to infinity. Since adding five on
8:51:30
the outside affects the y values, and the domain is the x values, this transformation doesn't change the domain. So the domain is still from zero to infinity. Now the range of our original y equals ln x was from negative infinity to infinity. Shifting up by five does affect the y values, and the range is talking about the y values. But since the original range was all real numbers, if you add five to all set of all real numbers, you
8:51:35
still get the set of all real numbers. So in this case, the range doesn't change either. And finally, we already saw that the original vertical asymptote of the y axis x equals zero, when we shift that up by five units, it's still the vertical line x equals zero. In this next example, we're starting with a log base 10 function. And since the plus two is on the inside, that means we shift that graph left by two. So I'll draw our basic
8:51:40
log function. Here's our basic log function. So I'll think of that as y equals log of x going through the point one, zero, here's its vertical asymptote. Now I need to shift everything left by two. So my vertical asymptote shifts left, and now it's at the line x equals negative two, instead of at x equals zero, and my graph, let's see my point, one zero gets shifted to, let's see negative one zero, since I'm subtracting two from the axis, and
8:51:45
here's a rough sketch of the resulting graph. Let's compare the features of the two graphs drawn here. We're talking about domains, the original kind of domain of from zero to infinity. But now I've shifted that left. So I subtracted two from all my exercises. And here's my new domain, which I can also verify just by looking at the picture. My range was originally from negative infinity to infinity. Well, shifting left only affects the x value, so it doesn't
8:51:50
even affect the range. So my range is still negative infinity to infinity. My vertical asymptote was originally at x equals zero. And since I subtract two from all my x values, that shifts that to x equals negative two. In this last problem, I'm not going to worry about drawing this graph. I'll just use algebra to compute its domain. So let's think about What's the issue, when you're taking the logs of things? Well, you can't take the log of
8:51:55
a negative number or zero. So whatever is inside the argument of the log function, whatever is being fed into log had better be greater than zero. So I'll write that down, we need to minus 3x, to be greater than zero. Now it's a matter of solving an inequality. Two has got to be greater than 3x. So two thirds is greater than x. In other words, x has to be less than two thirds. So our domain is all the x values from negative infinity to
8:52:00
two thirds, not including two thirds, it's a good idea to memorize the basic shape of the graph of a log function. It looks something like this, go through the point one zero, and has a vertical asymptote on the y axis. Also, if you remember that you can't take the log of a negative number, or zero, then that helps you quickly compute domains for log functions. Whatever's inside the log function, you set that greater than zero, and solve.
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This video is about combining logs and exponents. Please pause the video and take a moment to use your calculator to evaluate the following four expressions. Remember, that log base 10 on your calculator is the log button. While log base e on your calculator is the natural log button, you should find that the log base 10 of 10 cubed is three, the log base e of e to the 4.2 is 4.2 10 to the log base 10 of 1000 is 1000.
8:52:10
And eat the log base e of 9.6 is 9.6. In each case, the log and the exponential function with the same base undo each other, and we're left with the exponent. In fact, it's true that for any base a the log base a of a to the x is equal to x. The same sort of cancellation happens if we do the exponential function in the log function with the same base in the opposite order. For example, we take 10 to the power of log base 10 of 1000, the 10
8:52:15
to the power and the log base 10 undo each other, and we're left with the 1000s. This happens for any base a a to the log base a of x is equal to x. We can describe this by saying that an exponential function and a log function with the same base undo each other. If you're familiar with the language of inverse functions, the exponential function and log function are inverses. Let's see why these roles hold for the first log role. log
8:52:20
base a of a dx is asking the question, What power do we raise a two in order to get a to the x? In other words, a to what power is a to the x? Well, the answer is clearly x. And that's why log base a of a to the x equals x. For the second log rule, notice that the log base a of x means the power we raise a two to get x. But this expression is saying that we're supposed to raise a to that power. If we raise a to the power, we
8:52:25
need to raise a two to get x, then we'll certainly get x. Now let's use these two roles. In some examples. If we want to find three to the log base three of 1.43 to the power and log base three undo each other, so we're left with 1.4. Remember, that log base 10 on your calculator is the log button. While log base e on your calculator is the natural log button, you should find that the log base 10 of 10 cubed is three, the log base e of e to the 4.2 is 4.2 10 to the log base 10 of 1000 is 1000.
8:52:31
And eat the log base e of 9.6 is 9.6. In each case, the log and the exponential function with the same base undo each other, and we're left with the exponent. In fact, it's true that for any base a the log base a of a to the x is equal to x. The same sort of cancellation happens if we do the exponential function in the log function with the same base in the opposite order. For example, we take 10 to the power of log base 10 of 1000, the 10
8:52:36
to the power and the log base 10 undo each other, and we're left with the 1000s. This happens for any base a a to the log base a of x is equal to x. We can describe this by saying that an exponential function and a log function with the same base undo each other. If you're familiar with the language of inverse functions, the exponential function and log function are inverses. Let's see why these roles hold for the first log role. log
8:52:41
base a of a dx is asking the question, What power do we raise a two in order to get a to the x? In other words, a to what power is a to the x? Well, the answer is clearly x. And that's why log base a of a to the x equals x. For the second log rule, notice that the log base a of x means the power we raise a two to get x. But this expression is saying that we're supposed to raise a to that power. If we raise a to the power, we
8:52:46
need to raise a two to get x, then we'll certainly get x. Now let's use these two roles. In some examples. If we want to find three to the log base three of 1.43 to the power and log base three undo each other, so we're left with 1.4. If we want to find ln of e to the x, remember that ln means log base e, so we're taking log base e of e to the x. If we want to find ln of e to the x, remember that ln means log base e, so we're taking log base e of e to the x.
8:52:53
Well, those functions undo each other and we're left with x. If we want to take 10 to the log of three z, remember that log without a base written implies that the base is 10. So really, we want to take 10 to the log base 10 of three z will tend to a power and log base 10 undo each other. So we're left with a three z. Finally, does this last statement hold is ln of 10 to the x equal to x? Well, ln means log base e. So we're taking log base
8:52:58
e of 10 to the x, notice that the base of the log and the base of the exponential function are not the same thing. So they don't undo each other. And in fact, log base e of 10 to the x is not usually equal to x, we can check with one example, say if x equals one, then log base e of 10 to the one, that's log base e of 10. And we can check on the calculator that's equal to 2.3. In some more decimals, which is not the same thing as one. So this
8:53:03
statement is false, it does not hold. We need the basis to be the same for logs and exponent to undo each other. In this video, we saw that logs and exponents with the same base undo each other. Specifically, log base a of a to the x is equal to x and a to the log base a of x is also equal to x Well, those functions undo each other and we're left with x. If we want to take 10 to the log of three z, remember that log without a base written implies that the base is 10.
8:53:09
So really, we want to take 10 to the log base 10 of three z will tend to a power and log base 10 undo each other. So we're left with a three z. Finally, does this last statement hold is ln of 10 to the x equal to x? Well, ln means log base e. So we're taking log base e of 10 to the x, notice that the base of the log and the base of the exponential function are not the same thing. So they don't undo each other. And in fact, log base e of 10
8:53:14
to the x is not usually equal to x, we can check with one example, say if x equals one, then log base e of 10 to the one, that's log base e of 10. And we can check on the calculator that's equal to 2.3. In some more decimals, which is not the same thing as one. So this statement is false, it does not hold. We need the basis to be the same for logs and exponent to undo each other. In this video, we saw that logs and exponents with the same base
8:53:19
undo each other. Specifically, log base a of a to the x is equal to x and a to the log base a of x is also equal to x for any values of x and any base a. This video is about rules or properties of logs. The log rules are closely related to the exponent rules. So let's start by reviewing some of the exponent rules. for any values of x and any base a. This video is about rules or properties of logs. The log rules are closely related to the exponent rules. So let's start by reviewing some of
8:53:26
the exponent rules. To keep things simple, we'll write everything down with a base of two. Even though the exponent rules hold for any base, we know that if we raise two to the zero power, we get one, we have a product rule for exponents, which says that two to the M times two to the n is equal to two to the m plus n. In other words, if we multiply two numbers, then we add the exponents. We also have a quotient rule that says that two to the M divided by n to the n is equal
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to two to the m minus n. In words, that says that if we divide two numbers, then we subtract the exponents. Finally, we have a power rule that says if we take a power to a power, then we multiply the exponents. Each of these exponent rules can be rewritten as a log rule. The first rule, two to the zero equals one can be rewritten in terms of logs as log base two of one equals zero. That's because log base two of one equals zero means two to the
8:53:37
zero equals one. The second rule, the product rule, can be rewritten in terms of logs by saying log of x times y equals log of x plus log of y. I'll make these base two to agree with my base that I'm using for my exponent rules. In words, that says the log of the product is the sum of the logs. Since logs really represent exponents. This is saying that when you multiply two numbers together, you add their exponents, which is just what
8:53:42
we said for the exponent version. The quotient rule for exponents can be rewritten in terms of logs by saying the log of x divided by y is equal to the log of x minus the log of y. In words, we can say that the log of the quotient is equal to the difference of the logs. Since logs are really exponents, another way of saying the same thing is that when you divide two numbers, you subtract their exponents. That's how we described the exponent
8:53:47
rule above. Finally, the power rule for exponents can be rewritten in terms of logs by saying the log of x to the n is equal to n times log of x. Sometimes people describe this rule by saying when you take the log of an expression with an exponent, you can bring down the exponent and multiply. If we think of x as being some power of two, this is really saying when we take a power to a power, we multiply their exponents. That's exactly how we described
8:53:52
the power rule above. It doesn't really matter if you multiply this exponent on the left side, or on the right side, but it's more traditional to multiply it on the left side. I've given these rules with the base of two, but they actually work for any base. To help you remember them, please take a moment to write out the log rules using a base of a you should get the following chart. Let's use the log rules to rewrite the following
8:53:57
expressions as a sum or difference of logs. In the first expression, we have a log base 10 of a quotient. So we can rewrite the log of the quotient To keep things simple, we'll write everything down with a base of two. Even though the exponent rules hold for any base, we know that if we raise two to the zero power, we get one, we have a product rule for exponents, which says that two to the M times two to the n is equal
8:54:02
to two to the m plus n. In other words, if we multiply two numbers, then we add the exponents. We also have a quotient rule that says that two to the M divided by n to the n is equal to two to the m minus n. In words, that says that if we divide two numbers, then we subtract the exponents. Finally, we have a power rule that says if we take a power to a power, then we multiply the exponents. Each of these exponent rules can be rewritten as a log rule. The
8:54:07
first rule, two to the zero equals one can be rewritten in terms of logs as log base two of one equals zero. That's because log base two of one equals zero means two to the zero equals one. The second rule, the product rule, can be rewritten in terms of logs by saying log of x times y equals log of x plus log of y. I'll make these base two to agree with my base that I'm using for my exponent rules. In words, that says the log of the
8:54:12
product is the sum of the logs. Since logs really represent exponents. This is saying that when you multiply two numbers together, you add their exponents, which is just what we said for the exponent version. The quotient rule for exponents can be rewritten in terms of logs by saying the log of x divided by y is equal to the log of x minus the log of y. In words, we can say that the log of the quotient is equal to the difference of the
8:54:17
logs. Since logs are really exponents, another way of saying the same thing is that when you divide two numbers, you subtract their exponents. That's how we described the exponent rule above. Finally, the power rule for exponents can be rewritten in terms of logs by saying the log of x to the n is equal to n times log of x. Sometimes people describe this rule by saying when you take the log of an expression with an exponent, you can bring down the exponent
8:54:22
and multiply. If we think of x as being some power of two, this is really saying when we take a power to a power, we multiply their exponents. That's exactly how we described the power rule above. It doesn't really matter if you multiply this exponent on the left side, or on the right side, but it's more traditional to multiply it on the left side. I've given these rules with the base of two, but they actually work for any base. To help
8:54:27
you remember them, please take a moment to write out the log rules using a base of a you should get the following chart. Let's use the log rules to rewrite the following expressions as a sum or difference of logs. In the first expression, we have a log base 10 of a quotient. So we can rewrite the log of the quotient as the difference of the logs. Now we still have the log of a product, I can rewrite the log of a product as the sum of the logs
8:54:33
as the difference of the logs. Now we still have the log of a product, I can rewrite the log of a product as the sum of the logs So that is log of y plus log of z. When I put things together, I have to be careful, because here I'm subtracting the entire log expression. So I need to subtract both terms of this son. So that is log of y plus log of z. When I put things together, I have to be careful, because here I'm subtracting the entire log expression. So I need to subtract both terms
8:54:40
of this son. I'll make sure I do that by putting them in parentheses. Now I can simplify a little bit by distributing the negative sign. And here's my final answer. In my next expression, I have the log of a product. So I can rewrite that as the sum of two logs. I can also use my power rule to bring down the exponent T, and multiply it in the front. That gives me the final expression log of five plus I'll make sure I do that by putting them in parentheses. Now I can simplify a little bit
8:54:47
by distributing the negative sign. And here's my final answer. In my next expression, I have the log of a product. So I can rewrite that as the sum of two logs. I can also use my power rule to bring down the exponent T, and multiply it in the front. That gives me the final expression log of five plus t times log of two. One common mistake on this problem is to rewrite this expression as t times log of five times two. In fact, those two expressions are not equal. Because
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the T only applies to the two, not to the whole five times two, we can't just bring it down in front using the power rule. After all, the power rule only applies to a single expression raised to an exponent, and not to a product like this. And these next examples, we're going to go the other direction. Here, we're given sums and differences of logs. And we want to wrap them up into a single log expression. By look at the first two pieces,
8:54:58
that's a difference of logs. So I know I can rewrite it as the log of a quotient. Now I have the sum of two logs. So I can rewrite that as the log of a product. I'll clean that up a little bit and rewrite it as log base five of a times c over B. In my second example, I can rewrite the sum of my logs as the log of a product. Now, I would like to rewrite this difference of logs as the log of a quotient, but I can't do it yet, because of that factor
8:55:03
of two multiplied in front. But I can use the power rule backwards to put that two back up in the exponent. So I'll do that first. So I will copy down the ln of x plus one times x minus one, and rewrite this second term as ln of x squared minus one squared. Now I have a straightforward difference of two logs, which I can rewrite as the log of a quotient. I can actually simplify this some more. Since x plus one times x minus one is
8:55:08
the same thing as x squared minus one. I can cancel factors to get ln of one over x squared minus one. In this video, we solve for rules for logs that are related to exponent rules. First, we saw that the log with any base of one is equal to zero. Second, we saw the product rule, the log of a product is equal to the sum of the logs. We saw the quotient rule, the log of a quotient is the difference of the logs. And we saw the power rule. When
8:55:13
you take a log of an expression with an exponent in it, you can bring down the exponent and multiply it. It's worth noticing that there's no log rule that helps you split up the log of a song. In particular, the log of a psalm is not equal to the sum of the logs. If you think about logs and exponent rules going together, this kind of makes sense, because there's also no rule for rewriting the sum of two exponential expressions. Log roles
8:55:18
will be super handy. As we start to solve equations using locks. The chain rule is a really useful method for finding the derivative of the composition of two functions. Let's start with a brief review of composition. f composed with g means that we apply f to the output of G as a diagram, this means we start with x and apply g first. t times log of two. One common mistake on this problem is to rewrite this expression
8:55:23
as t times log of five times two. In fact, those two expressions are not equal. Because the T only applies to the two, not to the whole five times two, we can't just bring it down in front using the power rule. After all, the power rule only applies to a single expression raised to an exponent, and not to a product like this. And these next examples, we're going to go the other direction. Here, we're given sums and differences of logs.
8:55:28
And we want to wrap them up into a single log expression. By look at the first two pieces, that's a difference of logs. So I know I can rewrite it as the log of a quotient. Now I have the sum of two logs. So I can rewrite that as the log of a product. I'll clean that up a little bit and rewrite it as log base five of a times c over B. In my second example, I can rewrite the sum of my logs as the log of a product. Now, I would like to rewrite
8:55:33
this difference of logs as the log of a quotient, but I can't do it yet, because of that factor of two multiplied in front. But I can use the power rule backwards to put that two back up in the exponent. So I'll do that first. So I will copy down the ln of x plus one times x minus one, and rewrite this second term as ln of x squared minus one squared. Now I have a straightforward difference of two logs, which I can rewrite as the log of a
8:55:38
quotient. I can actually simplify this some more. Since x plus one times x minus one is the same thing as x squared minus one. I can cancel factors to get ln of one over x squared minus one. In this video, we solve for rules for logs that are related to exponent rules. First, we saw that the log with any base of one is equal to zero. Second, we saw the product rule, the log of a product is equal to the sum of the logs. We saw the quotient rule,
8:55:43
the log of a quotient is the difference of the logs. And we saw the power rule. When you take a log of an expression with an exponent in it, you can bring down the exponent and multiply it. It's worth noticing that there's no log rule that helps you split up the log of a song. In particular, the log of a psalm is not equal to the sum of the logs. If you think about logs and exponent rules going together, this kind of makes sense, because
8:55:48
there's also no rule for rewriting the sum of two exponential expressions. Log roles will be super handy. As we start to solve equations using locks. The chain rule is a really useful method for finding the derivative of the composition of two functions. Let's start with a brief review of composition. f composed with g means that we apply f to the output of G as a diagram, this means we start with x and apply g first.
8:55:53
Then we apply f to the output to get our final results I'm going to call G, the inner function, and F the outer function. Because g looks like it's on the inside of f, Then we apply f to the output to get our final results I'm going to call G, the inner function, and F the outer function. Because g looks like it's on the inside of f, in this standard notation, we can write h of x, which is the square root of sine of
8:55:58
x as the composition of two functions, by letting sine of x be the inner function, and the square root function be the outer function, which I write as f of u equals the square root of u. I like to do this sort of dissection of functions, by drawing a box around part of the function, whatever is inside the box becomes my inner function, whatever we do to the box becomes our outer function, in this case, taking the square root. This allows
8:56:03
us to write h of x as the composition, f of g of x, where f and g are the outer and inner functions defined here. Please take a moment to write the next two functions as compositions of functions, before you go on. A natural way to write k of x as a composition is to let our inner function be tan of x plus seacon of x. The outer function describes what happens to that box the inner function, it gets cubed and multiplied by five. There's several ways
8:56:08
to write the next example as a composition of functions. For example, we could take x squared as our inner function, and then our outer function takes a to the sign of that inner function. Alternatively, we could take the inner function to be sine of x squared. And then the outer function in this standard notation, we can write h of x, which is the square root of sine of x as the composition of two functions, by letting sine of x be the inner function, and
8:56:14
the square root function be the outer function, which I write as f of u equals the square root of u. I like to do this sort of dissection of functions, by drawing a box around part of the function, whatever is inside the box becomes my inner function, whatever we do to the box becomes our outer function, in this case, taking the square root. This allows us to write h of x as the composition, f of g of x, where f and g are the outer and inner
8:56:19
functions defined here. Please take a moment to write the next two functions as compositions of functions, before you go on. A natural way to write k of x as a composition is to let our inner function be tan of x plus seacon of x. The outer function describes what happens to that box the inner function, it gets cubed and multiplied by five. There's several ways to write the next example as a composition of functions. For example, we could take x
8:56:24
squared as our inner function, and then our outer function takes a to the sign of that inner function. Alternatively, we could take the inner function to be sine of x squared. And then the outer function has to be e to the power. It's also possible to write our function r of x as a composition of three functions. An inner function of x squared, a middle function of sine and an outermost function of e to the power, which are right as h of V equals e to the V.
8:56:30
has to be e to the power. It's also possible to write our function r of x as a composition of three functions. An inner function of x squared, a middle function of sine and an outermost function of e to the power, which are right as h of V equals e to the V. When calculating the derivatives of complicated functions, it's really important to recognize them as compositions of simpler functions. That way, we can build up the derivative in
8:56:35
terms of the simpler derivatives. And that's the idea behind the chain rule. The chain rule tells us if we have two differentiable functions, then the derivative of the composition f composed with g of x is equal to the derivative of the outer function evaluated on the inner function times the derivative of the inner function. Sometimes the chain rule is written instead, in lightness notation, that is the dydx notation. To see how this works, let's
8:56:40
let u equal g of x. And let's let y equal f of u. In other words, y is f of When calculating the derivatives of complicated functions, it's really important to recognize them as compositions of simpler functions. That way, we can build up the derivative in terms of the simpler derivatives. And that's the idea behind the chain rule. The chain rule tells us if we have two differentiable functions, then the derivative of the composition
8:56:45
f composed with g of x is equal to the derivative of the outer function evaluated on the inner function times the derivative of the inner function. Sometimes the chain rule is written instead, in lightness notation, that is the dydx notation. To see how this works, let's let u equal g of x. And let's let y equal f of u. In other words, y is f of g of x. g of x. Now do u dx is just another way of writing g prime of x, and d y d u is another way of
8:56:52
writing f prime of U. or in other words, f prime of g of x. Finally, if we write D y dX, that means we're taking the derivative of f composed with g. So that's f composed with g prime of x. Using this key, I can rewrite the expression above as the y dx equals d y d u Now do u dx is just another way of writing g prime of x, and d y d u is another way of writing f prime of U. or in other words, f prime of g of x. Finally, if we write D y
8:56:58
dX, that means we're taking the derivative of f composed with g. So that's f composed with g prime of x. Using this key, I can rewrite the expression above as the y dx equals d y d u times d u dx. These are the two alternative ways of writing the chain rule. Let's use the chain rule to take the derivative of the square root of sine x. times d u dx. These are the two alternative ways of writing the chain rule. Let's use the chain rule to take the derivative of the square root of sine x.
8:57:05
Actually, I'm going to rewrite this as h of x equals sine x to the one half power to make it easier to take derivatives. As a composition, we're thinking of the inner function as sine x and the outer function as the one half power. So the chain rule tells us that to take h prime Actually, I'm going to rewrite this as h of x equals sine x to the one half power to make it easier to take derivatives. As a composition, we're thinking of the inner function as sine
8:57:11
x and the outer function as the one half power. So the chain rule tells us that to take h prime of x we need to take the derivative of the outer function evaluated on the inner function and then multiply that by the derivative of the inner function, we know that the derivative of the inner function, sine x is just cosine x. And the derivative of the outer function is one half times u to the negative one half. So h prime of x is then one half times u to
8:57:17
the negative one half. But that's evaluated on the inner function, sine of of x we need to take the derivative of the outer function evaluated on the inner function and then multiply that by the derivative of the inner function, we know that the derivative of the inner function, sine x is just cosine x. And the derivative of the outer function is one half times u to the negative one half. So h prime of x is then one half times u to
8:57:22
the negative one half. But that's evaluated on the inner function, sine of x. x. And then we multiply that by cosine of x. Again, that's the derivative of the outer function evaluated on the inner function times the derivative of the inner function. And we found the derivative using the chain rule. For the next example, our inner function was tan x plus seacon X and our outer function, f of u was five u cubed. So k prime of x is
8:57:29
15 times u squared. But that's the evaluated on the inner function 10x plus seacon x, then we still need to multiply that by the derivative of the inner function 10x plus seacon x. So we get the 15 tan x plus secant x squared times the derivative of tan x, which is secant squared x, plus the derivative of secant x, which is secant x, tan x. And that's our chain rule derivative. In this last example, we're thinking of the outermost function as being
8:57:34
e to the power and its inner function is sine of x squared. But sine of x squared itself has an outer function of sine and an inner function of x squared. So to find r prime of x, we first have to take the derivative of the outermost function, well, the derivative of e to the power is just e to the power. And then we multiply that by cosine of x. Again, that's the derivative of the outer function evaluated on the inner function times the derivative of the inner function. And
8:57:40
we found the derivative using the chain rule. For the next example, our inner function was tan x plus seacon X and our outer function, f of u was five u cubed. So k prime of x is 15 times u squared. But that's the evaluated on the inner function 10x plus seacon x, then we still need to multiply that by the derivative of the inner function 10x plus seacon x. So we get the 15 tan x plus secant x squared times the derivative of tan x, which is secant
8:57:45
squared x, plus the derivative of secant x, which is secant x, tan x. And that's our chain rule derivative. In this last example, we're thinking of the outermost function as being e to the power and its inner function is sine of x squared. But sine of x squared itself has an outer function of sine and an inner function of x squared. So to find r prime of x, we first have to take the derivative of the outermost function, well, the derivative of e to the power is just e to the power.
8:57:51
And now we evaluate that on its inner function, sine of x squared. But now by the chain rule, we have to multiply that by the derivative of the inner function, sine of x squared. I'll copy down the E to the sine x squared. And I'll use the chain rule a second time to find the derivative of sine x squared. Now the outer function is sine, and the derivative of sine is cosine. I need to evaluate it on its inner function of x squared, and then
8:57:56
multiply that by the derivative of the inner function. After copying things down, I just have to take the derivative of x squared, which is 2x And now we evaluate that on its inner function, sine of x squared. But now by the chain rule, we have to multiply that by the derivative of the inner function, sine of x squared. I'll copy down the E to the sine x squared. And I'll use the chain rule a second time to find the derivative of sine x squared. Now the outer function is sine, and the derivative
8:58:02
of sine is cosine. I need to evaluate it on its inner function of x squared, and then multiply that by the derivative of the inner function. After copying things down, I just have to take the derivative of x squared, which is 2x by the power. by the power. This video introduced the chain rule, which says that the derivative of f composed with g at x is equal to f prime at g of x times g prime at x, or equivalently, d y dx is equal
8:58:09
to d y d u times d u dx. This video gives some more examples and a justification of the chain rule, and also includes a handy formula for the derivative of a to the x with respect to x, where A is any positive number. In the next example, I want to show using the chain rule that the derivative of five to the x is equal to ln five times five to the x. First, I want to rewrite five to the x as easily ln five times x. And I can do
8:58:14
that because e to the ln five is equal to five. So e to the ln five to the x power is equal to five to the x. But e to the ln five to the x power using exponent rules is just e to the ln five times x. So if I want to take the derivative of five to the x, after rewriting it as e to the ln five times x, let me think of the inner function as being ln five times x. And I'm going to think of the outer function as being E to that power.
8:58:19
That's what I wanted to Make the derivative of. So by the chain rule, I can first take the derivative of the outer function, derivative of e to the power is just e to the power, and I evaluate it at its inner function. But then by the chain rule, I need to take the derivative of the inner function, well, the derivative of ln five times x is just the constant coefficient ln five. And that's my derivative. Now, I know that e to the ln five
8:58:24
times x is just five to the x. That's what we talked about before. So my final answer is five to the x times ln five, or I guess I can rewrite that in the other order. That is that there's nothing special about five in this example, I could have done this same process with any a base positive base a. So I'm going to write that as a general principle that the derivative of a to the x with respect to x is equal to ln a times a to the x. This
8:58:29
is a fact worth memorizing. I'll use this fact to compute the derivative of this complicated expression, sine of 5x times the square root of two to the cosine 5x plus one. To find dydx, I'll first use the product rule, since our expression is the product of two other expressions. So D y dX is the first expression times the derivative of the second expression, which I'll go ahead and write using an exponent instead of a square root sign, plus the derivative
8:58:34
of the first expression times the second expression. Now I'll need to use the chain rule to evaluate the derivative here. My outermost function is the function that raises everything to the one half power. So when I take the derivative, I can use the power rule, bring down the one half, right to the cosine 5x plus one to the negative one half. But then by the chain rule, I'm going to have to multiply by the derivative of the inner function, which is to to the
8:58:39
cosine 5x plus one, I'll just carry along the rest of my expression for now, what I want to take the derivative of two to the cosine 5x plus one, I'm going to have to use the chain rule again, thinking of my outer function as being two to the power plus one. So let me copy things down on the next line. Now taking the derivative, the derivative of one is just zero, so I'm really just taking the derivative of two to the cosine 5x. And by my formula, this is going to be ln f two times two
8:58:45
This video introduced the chain rule, which says that the derivative of f composed with g at x is equal to f prime at g of x times g prime at x, or equivalently, d y dx is equal to d y d u times d u dx. This video gives some more examples and a justification of the chain rule, and also includes a handy formula for the derivative of a to the x with respect to x, where A is any positive number. In the next example, I want to show using
8:58:50
the chain rule that the derivative of five to the x is equal to ln five times five to the x. First, I want to rewrite five to the x as easily ln five times x. And I can do that because e to the ln five is equal to five. So e to the ln five to the x power is equal to five to the x. But e to the ln five to the x power using exponent rules is just e to the ln five times x. So if I want to take the derivative of five to the x, after
8:58:55
rewriting it as e to the ln five times x, let me think of the inner function as being ln five times x. And I'm going to think of the outer function as being E to that power. That's what I wanted to Make the derivative of. So by the chain rule, I can first take the derivative of the outer function, derivative of e to the power is just e to the power, and I evaluate it at its inner function. But then by the chain rule, I need to take the
8:59:00
derivative of the inner function, well, the derivative of ln five times x is just the constant coefficient ln five. And that's my derivative. Now, I know that e to the ln five times x is just five to the x. That's what we talked about before. So my final answer is five to the x times ln five, or I guess I can rewrite that in the other order. That is that there's nothing special about five in this example, I could have done this same
8:59:05
process with any a base positive base a. So I'm going to write that as a general principle that the derivative of a to the x with respect to x is equal to ln a times a to the x. This is a fact worth memorizing. I'll use this fact to compute the derivative of this complicated expression, sine of 5x times the square root of two to the cosine 5x plus one. To find dydx, I'll first use the product rule, since our expression is the product of two other
8:59:10
expressions. So D y dX is the first expression times the derivative of the second expression, which I'll go ahead and write using an exponent instead of a square root sign, plus the derivative of the first expression times the second expression. Now I'll need to use the chain rule to evaluate the derivative here. My outermost function is the function that raises everything to the one half power. So when I take the derivative, I can use the power rule, bring down the one
8:59:15
half, right to the cosine 5x plus one to the negative one half. But then by the chain rule, I'm going to have to multiply by the derivative of the inner function, which is to to the cosine 5x plus one, I'll just carry along the rest of my expression for now, what I want to take the derivative of two to the cosine 5x plus one, I'm going to have to use the chain rule again, thinking of my outer function as being two to the power plus one.
8:59:20
So let me copy things down on the next line. Now taking the derivative, the derivative of one is just zero, so I'm really just taking the derivative of two to the cosine 5x. And by my formula, this is going to be ln f two times two to the power of cosine 5x. But of course, I have to use the chain rule to the power of cosine 5x. But of course, I have to use the chain rule and multiply by that the derivative of the inner function here, which is cosine 5x. Again,
8:59:26
I'm just going to carry the rest of the expression along with me for the ride for now. Now we're taking the derivative of cosine 5x, I think of cosine as the outer function. And five times x is the inner function. Similarly, down here, sign is the outer function, and 5x is the inner function. So I can complete my work by copying a lot of stuff down, and now taking the derivative of cosine, which is minus sine, evaluated at its inner function,
8:59:31
times the derivative of the inner function 5x, which is just five. And I'm going to add to that the derivative of sine of 5x. Well, the derivative of sine is cosine, evaluated on its inner function times the derivative of the inner function 5x, which is just five times the rest of the stuff. I'll do a modest amount of simplification, maybe bring the constants out and combine any terms that I can. And that's the end of that complicated
8:59:36
example. And the next example, we'll try to find the derivative of a composition at the value x equals one just based on a table of values. So the chain rule says that the derivative of f composed with g is just going to be f prime evaluated g of x times g prime evaluated x, but I want to do this whole process at x equals one. So that's just going to be f prime at g of one times g prime of one. Well, g of one is two. So I really want f prime
8:59:41
at two, and f prime at two is 10. And g prime at one is negative five. So my answer is negative 50. I'm not going to give a rigorous proof of the chain rule. But I would like to give a more informal explanation based on the limit definition of derivative. So I'm going to write the derivative of f composed with g evaluated a point A, as the limit as x goes to a of f composed with g of x minus f composed with g of A divided by x minus a. I'll rewrite
8:59:46
this slightly. And now we're going to multiply the top and the bottom by g of x minus g of a, that doesn't change the value of expression, provided that g of x minus g of A is not zero. That's the detail on sweeping under the rug here and why this is not a real proof, but just a more informal explanation. Now if I rearrange things, and rewrite the limit of the product as the product of the limits, my limit on the right here is just the derivative
8:59:51
of g. for the limit on the left, notice that as x goes to a, g of x has to go to g of a, since G is differentiable on there for continuous function. So I can rewrite this and letting say u be equal to g of x, I can rewrite this as the limit as u goes to g of a of f of u minus f of g of A over u minus g of a. Now my expression on the left is just another way of writing the derivative of f evaluated at g of a. And I've arrived at the expression
8:59:56
for the chain rule. Let me just emphasize again, this is just a pseudo proof, it's not quite airtight, because g of x minus g of a might be zero. In this video, we saw some more examples of the chain rule justification of it. And we saw that the derivative respect to x and multiply by that the derivative of the inner function here, which is cosine 5x. Again, I'm just going to carry the rest of the expression along with me for the ride for now. Now we're
9:00:02
taking the derivative of cosine 5x, I think of cosine as the outer function. And five times x is the inner function. Similarly, down here, sign is the outer function, and 5x is the inner function. So I can complete my work by copying a lot of stuff down, and now taking the derivative of cosine, which is minus sine, evaluated at its inner function, times the derivative of the inner function 5x, which is just five. And I'm going to add
9:00:07
to that the derivative of sine of 5x. Well, the derivative of sine is cosine, evaluated on its inner function times the derivative of the inner function 5x, which is just five times the rest of the stuff. I'll do a modest amount of simplification, maybe bring the constants out and combine any terms that I can. And that's the end of that complicated example. And the next example, we'll try to find the derivative of a composition at the
9:00:12
value x equals one just based on a table of values. So the chain rule says that the derivative of f composed with g is just going to be f prime evaluated g of x times g prime evaluated x, but I want to do this whole process at x equals one. So that's just going to be f prime at g of one times g prime of one. Well, g of one is two. So I really want f prime at two, and f prime at two is 10. And g prime at one is negative five. So my answer is negative
9:00:17
50. I'm not going to give a rigorous proof of the chain rule. But I would like to give a more informal explanation based on the limit definition of derivative. So I'm going to write the derivative of f composed with g evaluated a point A, as the limit as x goes to a of f composed with g of x minus f composed with g of A divided by x minus a. I'll rewrite this slightly. And now we're going to multiply the top and the bottom by g of x minus g of
9:00:22
a, that doesn't change the value of expression, provided that g of x minus g of A is not zero. That's the detail on sweeping under the rug here and why this is not a real proof, but just a more informal explanation. Now if I rearrange things, and rewrite the limit of the product as the product of the limits, my limit on the right here is just the derivative of g. for the limit on the left, notice that as x goes to a, g of x has to go to g of a,
9:00:27
since G is differentiable on there for continuous function. So I can rewrite this and letting say u be equal to g of x, I can rewrite this as the limit as u goes to g of a of f of u minus f of g of A over u minus g of a. Now my expression on the left is just another way of writing the derivative of f evaluated at g of a. And I've arrived at the expression for the chain rule. Let me just emphasize again, this is just a pseudo proof, it's not
9:00:32
quite airtight, because g of x minus g of a might be zero. In this video, we saw some more examples of the chain rule justification of it. And we saw that the derivative respect to x of A to the X is equal to ln of a times a to the x. This video gives an explanation for why the chain rule holds. of A to the X is equal to ln of a times a to the x. This video gives an explanation for why the chain rule holds. I'm not going to give a rigorous proof of the chain rule. But I would like to give a
9:00:40
more informal explanation based on the limit definition of derivative. I'm not going to give a rigorous proof of the chain rule. But I would like to give a more informal explanation based on the limit definition of derivative. So I'm going to write the derivative of f composed with g evaluated a point A, as the limit as x goes to a of f composed with g of x minus f composed with g of A divided by x minus a. I'll rewrite this slightly. And now we're going to multiply the top and
9:00:46
the bottom by g of x minus g of a, that doesn't change the value of expression, provided that g of x minus g of A is not zero. That's the detail on sweeping under the rug here and why this is not a real proof, but just a more informal explanation. Now if I rearrange things, and rewrite the limit of the product as the product of the limits, my limit on the right here is just the derivative of g. for the limit on the left, notice that as x goes to
9:00:51
a, g of x has to go to G evey, since G is differentiable on there for a continuous function, so I can rewrite this and letting say u be equal to g of x, I can rewrite this as the limit as u goes to g of a of f of u minus f of g of A So I'm going to write the derivative of f composed with g evaluated a point A, as the limit as x goes to a of f composed with g of x minus f composed with g of A divided by x minus a. I'll rewrite this slightly. And now we're going to multiply the top and
9:00:57
the bottom by g of x minus g of a, that doesn't change the value of expression, provided that g of x minus g of A is not zero. That's the detail on sweeping under the rug here and why this is not a real proof, but just a more informal explanation. Now if I rearrange things, and rewrite the limit of the product as the product of the limits, my limit on the right here is just the derivative of g. for the limit on the left, notice that as x goes to
9:01:02
a, g of x has to go to G evey, since G is differentiable on there for a continuous function, so I can rewrite this and letting say u be equal to g of x, I can rewrite this as the limit as u goes to g of a of f of u minus f of g of A over over u minus g of a. Now my expression on the left is just another way of writing the derivative of f evaluated at G Ave. And I've arrived at the expression for the chain rule. Let
9:01:09
me just emphasize again, this is just a pseudo proof it's not quite airtight because g of x minus g of a might be zero. That's all for the justification of the chain rule. For a complete proof. Please see the textbook implicit differentiation is a technique for finding the slopes of tangent lines for curves that are defined indirectly, and sometimes aren't even functions. So far, we've developed a lot of techniques for finding derivatives
9:01:14
of functions defined explicitly, in terms of the equation y equals something. In this section, we'll consider curves that are defined implicitly, in terms of any equation involving x's and y's. So the points on this curve are the values of x and y that satisfy this equation. As you can see, when you have implicitly defined curves, they are not necessarily functions. And in fact, they can not only violate the vertical line test, but they can cross themselves,
9:01:19
or be broken up into several pieces or look like really cool pictures like this flower. But small pieces of these curves do satisfy the vertical line test for small pieces, y is a function of x. And that allows us to use our calculus techniques, especially the chain rule, to compute derivatives for these implicitly defined curves. u minus g of a. Now my expression on the left is just another way of writing the derivative
9:01:24
of f evaluated at G Ave. And I've arrived at the expression for the chain rule. Let me just emphasize again, this is just a pseudo proof it's not quite airtight because g of x minus g of a might be zero. That's all for the justification of the chain rule. For a complete proof. Please see the textbook implicit differentiation is a technique for finding the slopes of tangent lines for curves that are defined indirectly, and sometimes aren't
9:01:29
even functions. So far, we've developed a lot of techniques for finding derivatives of functions defined explicitly, in terms of the equation y equals something. In this section, we'll consider curves that are defined implicitly, in terms of any equation involving x's and y's. So the points on this curve are the values of x and y that satisfy this equation. As you can see, when you have implicitly defined curves, they are not necessarily functions.
9:01:34
And in fact, they can not only violate the vertical line test, but they can cross themselves, or be broken up into several pieces or look like really cool pictures like this flower. But small pieces of these curves do satisfy the vertical line test for small pieces, y is a function of x. And that allows us to use our calculus techniques, especially the chain rule, to compute derivatives for these implicitly defined curves.
9:01:39
As usual, the derivative dy dx represents the slope of a tangent line. For our first example, let's find the equation of the tangent line for the lips 9x squared plus four y squared equals 25. drawn below at the point, one, two. From the picture, it looks like the slope of this tangent line should be about negative one, but let's use calculus to find it exactly. So there are at least two ways we could proceed. First, we could solve for y, and then use
9:01:44
the same techniques that we've been using. So if we solve for y, we get for y squared equals 25 minus 9x squared. So y squared is 25 minus 9x squared over four, which means that y is plus or minus the square root of 25 minus 9x squared over four, or in other words, plus or minus the square root of 25 minus 9x squared over two, the plus answer is giving us the top half of the ellipse. And the minus answer is giving us this bottom.
9:01:49
Since the point one, two is on the top part of the ellipse, let's focus on the positive version. And let's take the derivative. But first, let me rewrite one more time to put it in a slightly easier form, instead of dividing by two, I'm going to think of multiplying by the constant one half. And instead of taking the square root, I'm going to write that as an exponent of one half here. So now if I want to take the y dx, I can pull out the
9:01:54
constant of one half. And now I'll start using the chain rule where my outer function As usual, the derivative dy dx represents the slope of a tangent line. For our first example, let's find the equation of the tangent line for the lips 9x squared plus four y squared equals 25. drawn below at the point, one, two. From the picture, it looks like the slope of this tangent line should be about negative one, but let's use calculus to find it exactly.
9:01:59
So there are at least two ways we could proceed. First, we could solve for y, and then use the same techniques that we've been using. So if we solve for y, we get for y squared equals 25 minus 9x squared. So y squared is 25 minus 9x squared over four, which means that y is plus or minus the square root of 25 minus 9x squared over four, or in other words, plus or minus the square root of 25 minus 9x squared over two, the plus answer
9:02:04
is giving us the top half of the ellipse. And the minus answer is giving us this bottom. Since the point one, two is on the top part of the ellipse, let's focus on the positive version. And let's take the derivative. But first, let me rewrite one more time to put it in a slightly easier form, instead of dividing by two, I'm going to think of multiplying by the constant one half. And instead of taking the square root, I'm going to write that as
9:02:09
an exponent of one half here. So now if I want to take the y dx, I can pull out the constant of one half. And now I'll start using the chain rule where my outer function is is taking things to the one half power, and my inner function is 25 minus 9x squared. So I'll take the derivative of my outer function by bringing the one half down, taking the inner function to the negative one half. Now I multiply by the derivative of the inner
9:02:16
function, which is negative 18x. If I simplify a little bit, I get D y dX is negative 18x over four times 25 minus 9x squared to the one half power, or in other words, dy dx is negative 9x over two times the square root of 25 minus 9x squared. This formula only holds for the top half of the ellipse for the bottom half, we would need to use the negative. Now I want to evaluate the derivative at the point one, two, so I'm going to take
9:02:21
dydx. When x equals one, I get negative nine over two times the square root of 25 minus nine, which is negative nine eighths. Since I've found the slope of the tangent line, and I know that point one, two is a point on the tangent line, I can now use the point slope form to write down the equation of the tangent line. simplified this becomes y equals negative nine is x plus nine eights plus two, or y equals negative nine, it's x plus 25.
9:02:26
It's now that we've solved the problem once using a familiar method. Let's go back to the beginning and solve it again using a new method. method two is implicit differentiation. The idea is that I'm going to take the derivative with respect to x of both sides. If my equation without having to solve for y, I can rewrite the left side as nine times the derivative of x squared plus four times the derivative of y squared. And the right side, the derivative
9:02:31
of a constant is zero. Going back to the left side, the derivative of x squared with respect to x is 2x. Now for the derivative of y squared with respect to x, I'm going to need to use the chain rule, I'm going to think of taking the squared power as my outside function, I'm going to think of y itself as my inside function, my inside function of x. Even though my entire curve is not a function, for small pieces of it, y is a function of x, so I can
9:02:36
get away with doing this, the derivative of my outside function y squared is to y, and the derivative of my inside function, y as a function of x is just dydx. Now I can solve for dy dx, which is going to tell me the slope of my tangent line. And so I get negative 18x from here, divided by eight y from here, which simplifies to negative nine fourths times X over Y. Notice that the formula for my derivative dydx has both x's and y's in
9:02:41
it. Of course, for this problem, if I wanted to, I could solve for y in terms of x using the original equation like I did in method one, and plug that in for y and get an expression entirely in terms of x, which should be the same as the expression I got previously. taking things to the one half power, and my inner function is 25 minus 9x squared. So I'll take the derivative of my outer function by bringing the one half down, taking the
9:02:46
inner function to the negative one half. Now I multiply by the derivative of the inner function, which is negative 18x. If I simplify a little bit, I get D y dX is negative 18x over four times 25 minus 9x squared to the one half power, or in other words, dy dx is negative 9x over two times the square root of 25 minus 9x squared. This formula only holds for the top half of the ellipse for the bottom half, we would need to use the
9:02:51
negative. Now I want to evaluate the derivative at the point one, two, so I'm going to take dydx. When x equals one, I get negative nine over two times the square root of 25 minus nine, which is negative nine eighths. Since I've found the slope of the tangent line, and I know that point one, two is a point on the tangent line, I can now use the point slope form to write down the equation of the tangent line. simplified this becomes y equals
9:02:56
negative nine is x plus nine eights plus two, or y equals negative nine, it's x plus 25. It's now that we've solved the problem once using a familiar method. Let's go back to the beginning and solve it again using a new method. method two is implicit differentiation. The idea is that I'm going to take the derivative with respect to x of both sides. If my equation without having to solve for y, I can rewrite the left side as nine times the derivative
9:03:01
of x squared plus four times the derivative of y squared. And the right side, the derivative of a constant is zero. Going back to the left side, the derivative of x squared with respect to x is 2x. Now for the derivative of y squared with respect to x, I'm going to need to use the chain rule, I'm going to think of taking the squared power as my outside function, I'm going to think of y itself as my inside function, my inside function of x. Even though
9:03:06
my entire curve is not a function, for small pieces of it, y is a function of x, so I can get away with doing this, the derivative of my outside function y squared is to y, and the derivative of my inside function, y as a function of x is just dydx. Now I can solve for dy dx, which is going to tell me the slope of my tangent line. And so I get negative 18x from here, divided by eight y from here, which simplifies to negative nine fourths
9:03:11
times X over Y. Notice that the formula for my derivative dydx has both x's and y's in it. Of course, for this problem, if I wanted to, I could solve for y in terms of x using the original equation like I did in method one, and plug that in for y and get an expression entirely in terms of x, which should be the same as the expression I got previously. But I don't really But I don't really need to do that in order to solve this problem. Instead, I can just plug in the x value of
9:03:18
one and the y value of two to get dy dx at x equals one, equal to negative nine fourths, times one half or negative nine, eight, which we'll recognize as the same answer we got before. So as before, we can compute the equation for the tangent line. And we'll again get y equals negative nine 8x plus 25, eights. In this example, implicit differentiation was a convenient way to find the derivative. But it was possible to solve for y and use
9:03:23
standard methods instead. But in many examples, like the next one, it's not possible to solve for y directly. And so implicit differentiation is the only way to go. implicit differentiation is definitely the key to finding y prime for this curve defined implicitly. So again, the idea is to take the derivative of both sides with respect to x, I can break this up into pieces. And now use the product rule for the first piece. So I get the first function x
9:03:28
cubed times the derivative of the second function y squared, the derivative of y squared is to y, d y dx, don't forget the dydx there, because y is a function of x plus the derivative of the first part 3x squared times the second part need to do that in order to solve this problem. Instead, I can just plug in the x value of one and the y value of two to get dy dx at x equals one, equal to negative nine fourths,
9:03:33
times one half or negative nine, eight, which we'll recognize as the same answer we got before. So as before, we can compute the equation for the tangent line. And we'll again get y equals negative nine 8x plus 25, eights. In this example, implicit differentiation was a convenient way to find the derivative. But it was possible to solve for y and use standard methods instead. But in many examples, like the next one, it's not possible to solve
9:03:38
for y directly. And so implicit differentiation is the only way to go. implicit differentiation is definitely the key to finding y prime for this curve defined implicitly. So again, the idea is to take the derivative of both sides with respect to x, I can break this up into pieces. And now use the product rule for the first piece. So I get the first function x cubed times the derivative of the second function y squared, the derivative of y squared is
9:03:43
to y, d y dx, don't forget the dydx there, because y is a function of x plus the derivative of the first part 3x squared times the second part y squared. y squared. Next, I need to take the derivative of sine x, y on the do use the chain rule here. So the derivative of the outside sine is cosine. And now I need to take the derivative of the inside x times y. And that's going to be a product rule application. So x times D y dX,
9:03:50
plus the derivative of x, which is just one times y. That all was just my left hand side. But fortunately, my right hand side is easier. The derivative of x cubed with respect to x is 3x squared. And the derivative of y cubed with respect to x is three y squared, dy dx. Now I need to solve for the y dx. And since it's scattered all over the place in three different places, I'm first going to distribute out to free it from these parentheses, and
9:03:55
then I'll try to move all the dydx is to the left side. So distributing out I get, I get this expression. And now moving alternatives with dydx and add them to the left side and all terms without dydx mm to the right side. I'm going to get this expression here. Now I'm going to factor out the dy dx. I'm just using standard algebra techniques here. And finally, I can just divide both sides by all this mess, too. isolate the dydx. And I found
9:04:00
my derivative using implicit differentiation. This video talked about using implicit differentiation to find the slopes of tangent lines for curves defined implicitly, the main two steps, were first to take the derivative of both sides with respect to x. And then to solve for dy dx. This video is about finding the derivatives of exponential functions, we've already seen that the derivative of the exponential function, e to the x is just itself, e to the x. But
9:04:05
what's the derivative of an exponential function with a different base, like five to the x, one way to find the derivative of an exponential function like five to the X is to write five as e to a power. So five, is the same thing as e to the ln five, where ln is the natural log or the log base. See, this makes sense because ln five, which is the same thing as log base e of five, means the power that we raise E to to get five. So now if we take
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e to the ln five, that means we raise e to the power that we raise E to to get five? Well, when you raise E to that power, you get five. All right, if five is the same thing as e to the ln five, then that means if we take five to the x, that's the same thing as e to the ln five raised to the x power BI properties of exponents, when I take a power to a power, I multiply the exponents. so this can be written as e to the ln five
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times x. Now I want to take the derivative with respect to x of five to the x. So by my rewriting trick, that's the same thing as taking the derivative with respect to x of e to the ln five times x. Now we know how to calculate this using the chain rule, we can think of e to the power as our outside function, and ln five times x as our inside function. So now by the chain rule, I take the derivative of the outside function e to
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the power, and that's just gives me e to the power evaluated on the inside function. So I stick ln five times x as my inside function, and by the chain rule, I multiply that by the derivative of the inside function, ln five times x. Five is a constant. So let me copy over first part, the derivative of a constant times x is just the constant. Let me rewrite this a little bit. So e to the ln five times x is the same thing as e to
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the ln five to the x power, just like before, because the exponent rules say, when I take a power to a power, I multiply the exponent, and remember, is the LL five, it's just a fancy way of writing five. So I've got five to the x times ln five as the derivative with respect to x of five to the x. The same argument works not just for a base five exponential function, but for any base exponential function. So if I take the respect to x of A to the X for any number A, I'm going to get
9:04:31
Next, I need to take the derivative of sine x, y on the do use the chain rule here. So the derivative of the outside sine is cosine. And now I need to take the derivative of the inside x times y. And that's going to be a product rule application. So x times D y dX, plus the derivative of x, which is just one times y. That all was just my left hand side. But fortunately, my right hand side is easier. The derivative of x cubed with respect to
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x is 3x squared. And the derivative of y cubed with respect to x is three y squared, dy dx. Now I need to solve for the y dx. And since it's scattered all over the place in three different places, I'm first going to distribute out to free it from these parentheses, and then I'll try to move all the dydx is to the left side. So distributing out I get, I get this expression. And now moving alternatives with dydx and add them to the left side and
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all terms without dydx mm to the right side. I'm going to get this expression here. Now I'm going to factor out the dy dx. I'm just using standard algebra techniques here. And finally, I can just divide both sides by all this mess, too. isolate the dydx. And I found my derivative using implicit differentiation. This video talked about using implicit differentiation to find the slopes of tangent lines for curves defined implicitly, the main two steps, were
9:04:46
first to take the derivative of both sides with respect to x. And then to solve for dy dx. This video is about finding the derivatives of exponential functions, we've already seen that the derivative of the exponential function, e to the x is just itself, e to the x. But what's the derivative of an exponential function with a different base, like five to the x, one way to find the derivative of an exponential function like five to the X is to write five
9:04:51
as e to a power. So five, is the same thing as e to the ln five, where ln is the natural log or the log base. See, this makes sense because ln five, which is the same thing as log base e of five, means the power that we raise E to to get five. So now if we take e to the ln five, that means we raise e to the power that we raise E to to get five? Well, when you raise E to that power, you get five. All right, if five is the same thing
9:04:56
as e to the ln five, then that means if we take five to the x, that's the same thing as e to the ln five raised to the x power BI properties of exponents, when I take a power to a power, I multiply the exponents. so this can be written as e to the ln five times x. Now I want to take the derivative with respect to x of five to the x. So by my rewriting trick, that's the same thing as taking the derivative with respect to x
9:05:01
of e to the ln five times x. Now we know how to calculate this using the chain rule, we can think of e to the power as our outside function, and ln five times x as our inside function. So now by the chain rule, I take the derivative of the outside function e to the power, and that's just gives me e to the power evaluated on the inside function. So I stick ln five times x as my inside function, and by the chain rule, I multiply that by
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the derivative of the inside function, ln five times x. Five is a constant. So let me copy over first part, the derivative of a constant times x is just the constant. Let me rewrite this a little bit. So e to the ln five times x is the same thing as e to the ln five to the x power, just like before, because the exponent rules say, when I take a power to a power, I multiply the exponent, and remember, is the LL five, it's just a
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fancy way of writing five. So I've got five to the x times ln five as the derivative with respect to x of five to the x. The same argument works not just for a base five exponential function, but for any base exponential function. So if I take the respect to x of A to the X for any number A, I'm going to get a to the x times ln A. Now, you might be wondering, what if I use the same roll on our old favorite e to the x. So our base here is E. That means I should get e to the x times ln e? Wait a
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sec, ln E, that's log base e of E, that's asking what power? Do I raise II today get he? Well, the answer there is one. And so the derivative respect to x of e to the x by this new rule we have is e to the x, it agrees with our old rule. I want to draw your attention to the difference between two expressions. And the first expression, dy dx of a to the x, the variable that we're taking the derivative with respect to is in the exponent. So for
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this exponential function, or we use the derivative rule that we just found, dy dx of a dx is eight of the x times ln A. On the other hand, if we take respect to x of x to the A, where the variable x that we're taking the derivative with respect to is in the base, then we don't need this exponential rule. In fact, it doesn't even apply, although we have here is the power rule, right? dy dx of x cubed would be 3x squared dy dx of x to the seventh would be
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7x to the sixth and enjoy Add x of x to the a is just a times x to the A minus one by the power role. So it's important to pay attention to where the variable is when you're taking a derivative. In this video, we found that the derivative with respect to x of five to the x is given by ln five times five to the x. And in general, the derivative of respect to x of A to the X is going to be ln a times a to the x. This gives us a general formula
9:05:32
for the derivative of exponential functions. The main goal of this video is to figure out the derivatives of logarithmic functions, functions like y equals ln x, or y equals log base A x for any positive base a, I want to find the derivative of log base a of x. In other words, I want to find the derivative of y, where y is log base a of x. By the definition of logarithms, log base a of x equals y means that a to the Y power is equal to x. And that's
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useful because now I can take the derivative of both sides and use implicit differentiation. Recall of the derivative of a to the power is ln A times A to the power. But since why we're thinking of as a function of x, I have to multiply that by dou y dx by the chain rule. The right hand side here is just one. Solving for dydx, I get one over ln A times A to the Y. But since age the y is equal to x, I can rewrite that as the y dx equals one
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over ln A times x. So the derivative of log base a of x for any base a is one over ln of A times x. And in particular, the derivative of natural log of x is one over ln of E times x. But since ln of E is just one, that saying that the derivative of ln x is one over x, this is a very handy fact. And this more general derivative is also worth memorizing. While we're talking about the derivative of the natural log of x, let's look at the derivative
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of the natural log of the absolute value of x. The function y equals ln of absolute value of x is of course closely related to the function y equals ln of x, the difference being that the domain for ln x is just x values greater than zero, whereas the domain of ln of absolute value of x is all X's not equal to zero. The graphs are also related. When you look at the graph of y equals ln absolute value of x, it looks like you're seeing double since
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the absolute value of x is equal to x, when x is greater than or equal to zero and negative x when x is less than zero, ln of the absolute value of x is going to be equal to ln x when x is greater than or equal to zero, and ln of negative x when x is less than zero. If I consider the derivative of ln of absolute value of x, I can think of taking the derivative of each piece separately. We just saw that the derivative of ln x is one over x. So the
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derivative of ln of minus x is going to be one over minus x times the derivative of minus x, which is minus one by the chain rule. Notice that this second expression simplifies to one over x. So the derivative of ln of absolute value of x is equal to one over x, whether x is positive or negative. This formula will come in handy later when we start doing integrals. In this video, we found that the derivative of ln x is equal to one over x, kind of a
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nice derivative. And more generally, the derivative of log base a of x is one over ln A times x. We've seen previously that the derivative of x to a constant A is equal to a times x to the A minus one. This is the power rule. We've also seen that the derivative of a positive number a raised to the x power is equal to ln A times A to the X So we know how to take the derivative when the variable x is in the base, or when it's in the exponent. But what
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if the variables in both the base and the exponent, how do we take the derivative of x to the x. To differentiate functions like this, we'll need to use the technique of logarithmic differentiation. To find the derivative of x to the x, I'm going to set y equal to x to the x. Now we want to find dy dx. Since we don't know how to compute dydx directly, so let's take the natural log of both sides. taking the log is often a handy trick when
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you have a variable in the exponent that you don't know how to deal with, because the properties of logs allow us to bring that exponent down and multiply it. Now we have y implicitly defined in terms of x, so let's use implicit differentiation, we'll take the derivative of both sides with respect to x. And now we should have no trouble taking the derivatives because we've gotten rid of the awkward exponential expression. So the derivative on the left
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of ln y is one over y times dy dx. And the derivative on the right using the product rule is x times one over x plus one times ln x. a to the x times ln A. Now, you might be wondering, what if I use the same roll on our old favorite e to the x. So our base here is E. That means I should get e to the x times ln e? Wait a sec, ln E, that's log base e of E, that's asking what power? Do I raise II today get he? Well, the answer there is one. And so the derivative respect to x of e to the x
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by this new rule we have is e to the x, it agrees with our old rule. I want to draw your attention to the difference between two expressions. And the first expression, dy dx of a to the x, the variable that we're taking the derivative with respect to is in the exponent. So for this exponential function, or we use the derivative rule that we just found, dy dx of a dx is eight of the x times ln A. On the other hand, if we take respect to x of x to the A, where
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the variable x that we're taking the derivative with respect to is in the base, then we don't need this exponential rule. In fact, it doesn't even apply, although we have here is the power rule, right? dy dx of x cubed would be 3x squared dy dx of x to the seventh would be 7x to the sixth and enjoy Add x of x to the a is just a times x to the A minus one by the power role. So it's important to pay attention to where the variable is when you're taking
9:06:33
a derivative. In this video, we found that the derivative with respect to x of five to the x is given by ln five times five to the x. And in general, the derivative of respect to x of A to the X is going to be ln a times a to the x. This gives us a general formula for the derivative of exponential functions. The main goal of this video is to figure out the derivatives of logarithmic functions, functions like y equals ln x, or y equals
9:06:38
log base A x for any positive base a, I want to find the derivative of log base a of x. In other words, I want to find the derivative of y, where y is log base a of x. By the definition of logarithms, log base a of x equals y means that a to the Y power is equal to x. And that's useful because now I can take the derivative of both sides and use implicit differentiation. Recall of the derivative of a to the power is ln A times A to the power. But since why
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we're thinking of as a function of x, I have to multiply that by dou y dx by the chain rule. The right hand side here is just one. Solving for dydx, I get one over ln A times A to the Y. But since age the y is equal to x, I can rewrite that as the y dx equals one over ln A times x. So the derivative of log base a of x for any base a is one over ln of A times x. And in particular, the derivative of natural log of x is one over ln of E times
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x. But since ln of E is just one, that saying that the derivative of ln x is one over x, this is a very handy fact. And this more general derivative is also worth memorizing. While we're talking about the derivative of the natural log of x, let's look at the derivative of the natural log of the absolute value of x. The function y equals ln of absolute value of x is of course closely related to the function y equals ln of x, the difference being that
9:06:53
the domain for ln x is just x values greater than zero, whereas the domain of ln of absolute value of x is all X's not equal to zero. The graphs are also related. When you look at the graph of y equals ln absolute value of x, it looks like you're seeing double since the absolute value of x is equal to x, when x is greater than or equal to zero and negative x when x is less than zero, ln of the absolute value of x is going to be equal to ln x when
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x is greater than or equal to zero, and ln of negative x when x is less than zero. If I consider the derivative of ln of absolute value of x, I can think of taking the derivative of each piece separately. We just saw that the derivative of ln x is one over x. So the derivative of ln of minus x is going to be one over minus x times the derivative of minus x, which is minus one by the chain rule. Notice that this second expression simplifies to
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one over x. So the derivative of ln of absolute value of x is equal to one over x, whether x is positive or negative. This formula will come in handy later when we start doing integrals. In this video, we found that the derivative of ln x is equal to one over x, kind of a nice derivative. And more generally, the derivative of log base a of x is one over ln A times x. We've seen previously that the derivative of x to a constant A is equal to a times x
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to the A minus one. This is the power rule. We've also seen that the derivative of a positive number a raised to the x power is equal to ln A times A to the X So we know how to take the derivative when the variable x is in the base, or when it's in the exponent. But what if the variables in both the base and the exponent, how do we take the derivative of x to the x. To differentiate functions like this, we'll need to use the technique of logarithmic
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differentiation. To find the derivative of x to the x, I'm going to set y equal to x to the x. Now we want to find dy dx. Since we don't know how to compute dydx directly, so let's take the natural log of both sides. taking the log is often a handy trick when you have a variable in the exponent that you don't know how to deal with, because the properties of logs allow us to bring that exponent down and multiply it. Now we have y implicitly
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defined in terms of x, so let's use implicit differentiation, we'll take the derivative of both sides with respect to x. And now we should have no trouble taking the derivatives because we've gotten rid of the awkward exponential expression. So the derivative on the left of ln y is one over y times dy dx. And the derivative on the right using the product rule is x times one over x plus one times ln x. This simplifies to one over y d y dx equals one plus ln x. So D y dX is going to equal
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y times one plus ln x. and replacing y with x to the x, I have dy dx is x to the x times one plus ln x. This technique of taking the log of both sides differentiating and solving for dydx is known as logarithmic differentiation. And it's enormously useful whenever you have variables in both the base and the exponent. Here's another example where our variable is in both the base and the exponent. So as before, I'm going to set y equal to the expression
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that I want to differentiate and compute dydx. First, I'll take the log of both sides. Use my log rules to bring my exponent down and multiply it and take the derivative of both sides with respect to x. On the left, I get one over y d y dx. And on the right, I get one of our x times the derivative of ln tangent x, which is one over tangent x times the derivative of tangent x, or secant squared x, continuing with the product rule, and didn't take the
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derivative of one of our x, that's going to be the derivative of x to the minus one, which is minus one times x to the minus This simplifies to one over y d y dx equals one plus ln x. So D y dX is going to equal y times one plus ln x. and replacing y with x to the x, I have dy dx is x to the x times one plus ln x. This technique of taking the log of both sides differentiating and solving for dydx is known as logarithmic differentiation. And it's enormously useful whenever you have
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variables in both the base and the exponent. Here's another example where our variable is in both the base and the exponent. So as before, I'm going to set y equal to the expression that I want to differentiate and compute dydx. First, I'll take the log of both sides. Use my log rules to bring my exponent down and multiply it and take the derivative of both sides with respect to x. On the left, I get one over y d y dx. And on the right, I get
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one of our x times the derivative of ln tangent x, which is one over tangent x times the derivative of tangent x, or secant squared x, continuing with the product rule, and didn't take the derivative of one of our x, that's going to be the derivative of x to the minus one, which is minus one times x to the minus two two times ln tangent of x. Simplifying the right hand side, I get one over x times one over
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sine x over cosine x times one over cosine squared x minus ln 10x over x squared. rewriting, I can flip and multiply to get one over x times cosine x over sine x times one over cosine squared x minus the second term, canceling one copy of cosine and rewriting in terms of cosecant and secant, I get this expression, I still have to solve for dy dx. times ln tangent of x. Simplifying the right hand side, I get one over x times one over
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sine x over cosine x times one over cosine squared x minus ln 10x over x squared. rewriting, I can flip and multiply to get one over x times cosine x over sine x times one over cosine squared x minus the second term, canceling one copy of cosine and rewriting in terms of cosecant and secant, I get this expression, I still have to solve for dy dx. So multiplying both sides by y, I get the following. And since y was equal to 10x to
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the one over x, I can rewrite everything in terms of x. The technique of logarithmic differentiation is most useful when taking the derivative of an expression that has a variable in both the base and the exponent, like in this example, but sometimes it's also handy just as a way to take the derivative of a complicated product and quotient like in this example. Now, we could take the derivative here just by using the quotient rule on the product rule, but
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it's a little easier to take the log of both sides. And the reason is that when we take the log of a product, we get a sum and the log of a quotient is a difference and sums and quotients are a lot easier to deal with. So in this example, the log of y is equal to ln of x plus ln of cosine of x minus ln of x squared plus x to the fifth power, I can even bring that fifth power down, because that's another one of my log roles. Now, it's
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much more straightforward to take the log of both sides. On the left, I have one over y dydx, as usual, and on the right, the derivative of ln x is one over x, the derivative of ln cosine x is one over cosine of x times negative sine of x. And the derivative of ln x squared plus x is one over x squared plus x times 2x plus one. I'll solve for dydx and get y times one over x minus sine x over cosine x, that's the same as tangent x minus five
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times 2x plus one over x squared plus x. Now I can just rewrite y in terms of x and I'll be done. Again, I didn't have to use logarithmic differentiation. To find this derivative, I could have just used the product rule in the quotient rule, but logarithmic differentiation made it computationally much easier. In this video, we learned how to take the derivative of expressions that have a variable both in the base and in the exponent. And the idea
9:08:22
was first to set y equal to the expression we want to derive. Next, to take the natural log of both sides. Next, to derive both sides. And finally, to solve for dy dx. This process is called logarithmic differentiation. The inverse of a function undoes what the function does, so the inverse of tying your shoes would be to untie them. And the inverse of the function that adds two to a number would be the function that subtracts two from a number. This video
9:08:27
introduces inverses and their properties. Suppose f of x is the function defined by this chart. In other words, f of two is three, f of three is five, f of four is six, and f of five is one, the inverse function for F written f superscript. Negative 1x undoes what f does. Since f takes two to three, F inverse takes three back to two. So we write this f superscript, negative one of three is two. Similarly, since f takes three to
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five, F inverse takes five to three. And since f takes four to six, f inverse of six is four. And since f takes five to one, f inverse of one is five. I'll use these numbers to fill in the chart. Notice that the chart of values when y equals f of x and the chart of values when y equals f inverse of x are closely related. They share the same numbers, but the x values for f of x correspond to the y values for f inverse of x, and the y values for f of
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x correspond to the x values for f inverse of x. That leads us to the first key fact inverse functions reverse the roles of y and x. I'm going to plot the points for y equals f of x in blue. So multiplying both sides by y, I get the following. And since y was equal to 10x to the one over x, I can rewrite everything in terms of x. The technique of logarithmic differentiation is most useful when taking the derivative of an expression that has a variable in both
9:08:43
the base and the exponent, like in this example, but sometimes it's also handy just as a way to take the derivative of a complicated product and quotient like in this example. Now, we could take the derivative here just by using the quotient rule on the product rule, but it's a little easier to take the log of both sides. And the reason is that when we take the log of a product, we get a sum and the log of a quotient is a difference and sums
9:08:48
and quotients are a lot easier to deal with. So in this example, the log of y is equal to ln of x plus ln of cosine of x minus ln of x squared plus x to the fifth power, I can even bring that fifth power down, because that's another one of my log roles. Now, it's much more straightforward to take the log of both sides. On the left, I have one over y dydx, as usual, and on the right, the derivative of ln x is one over x, the derivative of ln
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cosine x is one over cosine of x times negative sine of x. And the derivative of ln x squared plus x is one over x squared plus x times 2x plus one. I'll solve for dydx and get y times one over x minus sine x over cosine x, that's the same as tangent x minus five times 2x plus one over x squared plus x. Now I can just rewrite y in terms of x and I'll be done. Again, I didn't have to use logarithmic differentiation. To find this derivative,
9:08:58
I could have just used the product rule in the quotient rule, but logarithmic differentiation made it computationally much easier. In this video, we learned how to take the derivative of expressions that have a variable both in the base and in the exponent. And the idea was first to set y equal to the expression we want to derive. Next, to take the natural log of both sides. Next, to derive both sides. And finally, to solve for dy dx. This process
9:09:03
is called logarithmic differentiation. The inverse of a function undoes what the function does, so the inverse of tying your shoes would be to untie them. And the inverse of the function that adds two to a number would be the function that subtracts two from a number. This video introduces inverses and their properties. Suppose f of x is the function defined by this chart. In other words, f of two is three, f of three is five, f of four is six, and
9:09:08
f of five is one, the inverse function for F written f superscript. Negative 1x undoes what f does. Since f takes two to three, F inverse takes three back to two. So we write this f superscript, negative one of three is two. Similarly, since f takes three to five, F inverse takes five to three. And since f takes four to six, f inverse of six is four. And since f takes five to one, f inverse of one is five. I'll use these numbers to fill
9:09:13
in the chart. Notice that the chart of values when y equals f of x and the chart of values when y equals f inverse of x are closely related. They share the same numbers, but the x values for f of x correspond to the y values for f inverse of x, and the y values for f of x correspond to the x values for f inverse of x. That leads us to the first key fact inverse functions reverse the roles of y and x. I'm going to plot the points for y equals f of x in blue.
9:09:19
Next, I'll plot the points for y equals f inverse of x in red. Pause the video for a moment and see what kind of symmetry you observe in this graph. How are the blue points related to the red points, you might have noticed that the blue points and the red points are mirror images over the mirror line, y equals x. So our second key fact is that the graph of y equals f inverse of x can be obtained from the graph of y equals f of x by reflecting
9:09:24
over the line y equals x. This makes sense, because inverses, reverse the roles of y and x. In the same example, let's compute f inverse of f of two. This open circle means composition. In other words, we're computing f inverse of f of two. We compute this from the inside out. So that's f inverse of three. Since F of two is three, and f inverse of three, we see is two. Similarly, we can compute f of f inverse of three. And that means we take
9:09:29
f of f inverse of three. Since f inverse of three is two, that's the same thing as computing F of two, which is three. Please pause the video for a moment and compute these other compositions. You should have found that in every case, if you take f inverse of f of a number, you get back to the very same number you started with. And similarly, if you take f of f inverse of any number, you get back to the same number you started with. So in
9:09:34
general, f inverse of f of x is equal to x, and f of f inverse of x is also equal to x. This is the mathematical way of saying that F and envir f inverse undo each other. Let's look at a different example. Suppose that f of x is x cubed. Pause the video for a moment and guess what the inverse of f should be? Remember, F inverse undoes the work that F does. You might have guessed that f inverse of x is going to be the cube root function.
9:09:39
And we can check that this is true by looking at f of f inverse of x, that's F of the cube root of function, which means the cube root function cubed, which gets us back to x. Similarly, if we compute f inverse of f of x, that's the cube root of x cubed. And we get back to excellence again. So the cube root function really is the inverse of the cubing function. When we compose the two functions, we get back to the number that we started with. It'd
9:09:44
be nice to have a more systematic way of finding inverses of functions besides guessing and checking. One method uses the fact that inverses, reverse the roles of y and x. So if we want to find the inverse of the function, f of x equals five minus x over 3x, we can write it as y equals five minus x over 3x. Reverse the roles of y and x to get x equals five minus y over three y, and then solve for y. To solve for y, let's multiply both sides
9:09:49
by three y. Bring all terms with wisened to the left side, and alternans without y's and then to the right side, factor out the y and divide to isolate y. This gives us f inverse of x as five over 3x plus one. Notice that our original function f and our inverse function, f inverse are both rational functions, but they're not the reciprocals of each other. And then General, f inverse of x is not usually equal to one over f of x. This can be confusing,
9:09:54
because when we write two to the minus one, that does mean one of our two, but f to the minus one of x means the inverse function and not the reciprocal. It's natural to ask if all functions have inverse functions, that is for any function you might encounter. Is there always a function that it is its inverse? Next, I'll plot the points for y equals f inverse of x in red. Pause the video for a moment and see what kind of symmetry you observe in this graph. How are the blue points related
9:10:00
to the red points, you might have noticed that the blue points and the red points are mirror images over the mirror line, y equals x. So our second key fact is that the graph of y equals f inverse of x can be obtained from the graph of y equals f of x by reflecting over the line y equals x. This makes sense, because inverses, reverse the roles of y and x. In the same example, let's compute f inverse of f of two. This open circle means composition.
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In other words, we're computing f inverse of f of two. We compute this from the inside out. So that's f inverse of three. Since F of two is three, and f inverse of three, we see is two. Similarly, we can compute f of f inverse of three. And that means we take f of f inverse of three. Since f inverse of three is two, that's the same thing as computing F of two, which is three. Please pause the video for a moment and compute these other
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compositions. You should have found that in every case, if you take f inverse of f of a number, you get back to the very same number you started with. And similarly, if you take f of f inverse of any number, you get back to the same number you started with. So in general, f inverse of f of x is equal to x, and f of f inverse of x is also equal to x. This is the mathematical way of saying that F and envir f inverse undo each other. Let's
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look at a different example. Suppose that f of x is x cubed. Pause the video for a moment and guess what the inverse of f should be? Remember, F inverse undoes the work that F does. You might have guessed that f inverse of x is going to be the cube root function. And we can check that this is true by looking at f of f inverse of x, that's F of the cube root of function, which means the cube root function cubed, which gets us back to x. Similarly,
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if we compute f inverse of f of x, that's the cube root of x cubed. And we get back to excellence again. So the cube root function really is the inverse of the cubing function. When we compose the two functions, we get back to the number that we started with. It'd be nice to have a more systematic way of finding inverses of functions besides guessing and checking. One method uses the fact that inverses, reverse the roles of y and x. So if we want
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to find the inverse of the function, f of x equals five minus x over 3x, we can write it as y equals five minus x over 3x. Reverse the roles of y and x to get x equals five minus y over three y, and then solve for y. To solve for y, let's multiply both sides by three y. Bring all terms with wisened to the left side, and alternans without y's and then to the right side, factor out the y and divide to isolate y. This gives us f inverse
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of x as five over 3x plus one. Notice that our original function f and our inverse function, f inverse are both rational functions, but they're not the reciprocals of each other. And then General, f inverse of x is not usually equal to one over f of x. This can be confusing, because when we write two to the minus one, that does mean one of our two, but f to the minus one of x means the inverse function and not the reciprocal. It's natural to ask
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if all functions have inverse functions, that is for any function you might encounter. Is there always a function that it is its inverse? In fact, the answer is no. See, if you can come up with an example of a function that does not have an inverse function. The word function here is key. Remember that a function is a relationship between x values and y values, such that for each x value in the domain, there's only one corresponding y value. One example of a function that does not have an
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inverse function is the function f of x equals x squared. To see that, the inverse of this function is not a function. Note that for the x squared function, the number two and the number negative two, both go to number four. So if I had an inverse, he would have to send four to both two and negative two. The inverse would not be a function, it might be easier to understand the problem, when you look at a graph of y equals x squared.
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Recall that inverse functions reverse the roles of y and x and flip the graph over the line y equals x. But when I flipped the green graph over the line y equals x, I get this red graph. This red graph is not the graph of a function, because it violates the vertical line test. The reason that violates the vertical line test is because the original green function violates the horizontal line test, and has 2x values with the same y value. In general,
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a function f has an inverse function if and only if the graph of f satisfies the horizontal line test, ie every horizontal line intersects the graph. In most one point, pause the video for a moment and see which of these four graphs satisfy the horizontal line test. In other words, which of the four corresponding functions would have an inverse function, you may have found that graphs A and B, violate the horizontal line test. So their functions would not have
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inverse functions. But graph C and D satisfy the horizontal line test. So these graphs represent functions that do have inverses. functions that satisfy the horizontal line test are sometimes called One to One functions. Equivalent way of function is one to one, if for any two different x values, x one and x two, the y value is f of x one and f of x two are different numbers. Sometimes, as I said, f is one to one, if, whenever f of
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x one is equal to f of x two, then x one has to equal x two. As our last example, let's try to find P inverse of x, where p of x is the square root of x minus two drawn here. If we graph P inverse on the same axis as p of x, we get the following graph, simply by flipping over the line y equals x. If we try to solve the problem, algebraically, we can write y equal to a squared of x minus two, reverse the roles of y and x and solve
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for y by squaring both sides and adding two. Now if we were to graph y equals x squared plus two, that would look like a parabola, it would look like the red graph was already drawn together with another arm on the left side. But we know that our actual inverse function consists only of this right arm, we can specify this algebraically by making the restriction that x has to be bigger than or equal to zero. This corresponds to the
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fact that on the original graph, for the square root of x, y was only greater than or equal to zero. Looking more closely at the domain and range of P and P inverse, we know that the domain of P is all values of x such that x minus two is greater than or equal to zero. Since we can't take the square root of a negative number. This corresponds to x values being greater than or equal to two, or an interval notation, the interval from two to infinity.
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The range of P, we can see from the graph is our y value is greater than or equal to zero, or the interval from zero to infinity. In fact, the answer is no. See, if you can come up with an example of a function that does not have an inverse function. The word function here is key. Remember that a function is a relationship between x values and y values, such that for each x value in the domain, there's only one corresponding y value. One example of a function that does not have an
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inverse function is the function f of x equals x squared. To see that, the inverse of this function is not a function. Note that for the x squared function, the number two and the number negative two, both go to number four. So if I had an inverse, he would have to send four to both two and negative two. The inverse would not be a function, it might be easier to understand the problem, when you look at a graph of y equals x squared.
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Recall that inverse functions reverse the roles of y and x and flip the graph over the line y equals x. But when I flipped the green graph over the line y equals x, I get this red graph. This red graph is not the graph of a function, because it violates the vertical line test. The reason that violates the vertical line test is because the original green function violates the horizontal line test, and has 2x values with the same y value. In general,
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a function f has an inverse function if and only if the graph of f satisfies the horizontal line test, ie every horizontal line intersects the graph. In most one point, pause the video for a moment and see which of these four graphs satisfy the horizontal line test. In other words, which of the four corresponding functions would have an inverse function, you may have found that graphs A and B, violate the horizontal line test. So their functions would not have
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inverse functions. But graph C and D satisfy the horizontal line test. So these graphs represent functions that do have inverses. functions that satisfy the horizontal line test are sometimes called One to One functions. Equivalent way of function is one to one, if for any two different x values, x one and x two, the y value is f of x one and f of x two are different numbers. Sometimes, as I said, f is one to one, if, whenever f of
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x one is equal to f of x two, then x one has to equal x two. As our last example, let's try to find P inverse of x, where p of x is the square root of x minus two drawn here. If we graph P inverse on the same axis as p of x, we get the following graph, simply by flipping over the line y equals x. If we try to solve the problem, algebraically, we can write y equal to a squared of x minus two, reverse the roles of y and x and solve
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for y by squaring both sides and adding two. Now if we were to graph y equals x squared plus two, that would look like a parabola, it would look like the red graph was already drawn together with another arm on the left side. But we know that our actual inverse function consists only of this right arm, we can specify this algebraically by making the restriction that x has to be bigger than or equal to zero. This corresponds to the
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fact that on the original graph, for the square root of x, y was only greater than or equal to zero. Looking more closely at the domain and range of P and P inverse, we know that the domain of P is all values of x such that x minus two is greater than or equal to zero. Since we can't take the square root of a negative number. This corresponds to x values being greater than or equal to two, or an interval notation, the interval from two to infinity.
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The range of P, we can see from the graph is our y value is greater than or equal to zero, or the interval from zero to infinity. Similarly, based on the graph, we see the domain of P inverse is x values greater than or equal to zero, the interval from zero to infinity. And the range of P inverse is Y values greater than or equal to two, or the interval from two to infinity. If you look closely at these domains and ranges, you'll notice that the domain of P corresponds exactly
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to the range of P inverse, and the range of P corresponds to the domain of P inverse. This makes sense because inverse functions reverse the roles of y and x. The domain of f inverse of x is the x values for F inverse, which corresponds to the y values or the range of F. The range of f inverse is the y values for F inverse, which correspond to the x values or the domain of f. In this video, we discussed Similarly, based on the graph, we see the domain of P inverse is x values greater than
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or equal to zero, the interval from zero to infinity. And the range of P inverse is Y values greater than or equal to two, or the interval from two to infinity. If you look closely at these domains and ranges, you'll notice that the domain of P corresponds exactly to the range of P inverse, and the range of P corresponds to the domain of P inverse. This makes sense because inverse functions reverse the roles of y and x. The domain of
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f inverse of x is the x values for F inverse, which corresponds to the y values or the range of F. The range of f inverse is the y values for F inverse, which correspond to the x values or the domain of f. In this video, we discussed five key properties of inverse functions. inverse functions reverse the roles of y and x. The graph of y equals f inverse of x is the graph of y equals f of x reflected over the line y equals x. When we compose F with F inverse, we get the identity function y
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equals x. And similarly, when we compose f inverse with F, that brings x to x. In other words, F and F inverse undo each other. The function f of x has an inverse function if and only if the graph of y equals f of x satisfies the horizontal line test. And finally, the domain of f is the range of f inverse. And the range of f is the domain of f inverse. These properties of inverse functions will be important when we study exponential functions
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and their inverses logarithmic functions. This video defines the standard inverse trig functions, sine inverse cosine inverse and tan inverse. In this crazy looking graph, please focus first on the thin black line. This is a graph of y equals sine x. The graph of the inverse of a function can be found by flipping the graph of the original function over the line y equals x. I've drawn the flipped graph with this blue dotted line. But you'll
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notice that the blue dotted line is not the graph of a function, because it violates the vertical line test. So in order to get a function, that's the inverse of y equals sine x, we need to restrict the domain of sine of x, we'll restrict it to this piece that's drawn with a thick black line. If I invert that piece, by flipping it over the line y equals x, I get the piece drawn with a red dotted line here. And that piece does satisfy the
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vertical line test. So it is in fact a function. of the regular sine x has domain from negative infinity to infinity, or restricted sine x has domain from negative pi over two to pi over two. It's it's range is still from negative one to one, just like the regular sine x. Because I've taken the biggest possible piece of the graph whose flipped version is still a function. The inverse sine function is often written as arc sine of x. And since inverting
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a function reverses the roles of y at x, it reverses the domain and the range. So arc sine of x, the inverse function has domain from negative one to one, and range from negative pi over two to pi over two, which seems plausible from the graph. Now an inverse function undoes the work of a function. So if the function sine takes angle theta, two numbers, x, then the inverse sine, or arc sine takes numbers x, two angles theta. For example, since sine
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of pi over two is one, arc sine of one is pi over two. And in general, the output of arc sine of x is the angle between negative pi over two and pi over two whose side is x. y is equal to arc sine x means that x is equal to sine of y. But since there are many angles, y who sine is x, right, they all differ by multiples of two pi. We specify also, that y is between negative pi over two and pi over two. That was the whole point of doing this
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domain restriction in order to get a well defined inverse value. There's an alternate notation for inverse sine. Sometimes it's written as sine to the negative one of x. But this notation can be confusing, so be careful. In particular, sine to the negative one of x does not equal one over sine of x. One over sine of x the reciprocal function is another word for cosecant The backs, but sign to the negative one of x is another word for arc sine of x, the inverse sine function,
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which is not the same thing as the reciprocal function. Let's go through the same process to build an inverse cosine function, we start with a graph of cosine of x, we flip it over the line y equals x to get the blue dotted line. But the blue dotted line is not a function. So we go back and restrict the domain for our original cosine of x to just be between zero and pi. The resulting red graph now satisfies the vertical line test. So it's a proper inverse
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function. Our restricted cosine has domain from zero to pi, and range from negative one to one. And so our inverse function, arc cosine has domain from negative one to one, and range from zero to pi. Since cosine takes us from angles to numbers, arc cosine takes us from numbers back to angles. For example, cosine of pi over four is the square root of two over two. So arc cosine of the square root of two over two is equal to pi over four.
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arc cosine of x is the angle between zero and pi, whose cosine is x. In other words, y equals r cosine of x means that x is equal to cosine of y, and y is between zero and pi. Since otherwise, there'd be lots of possible answers for an angle y whose cosine is x. The alternative notation for arc cosine is cosine inverse. And again, we have to be careful. cosine to the negative 1x is not the same thing as one over cosine of x. one over cosine
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of x is also called secant of x. cosine to the negative 1x means arc cosine, the inverse function, and these two things are not the same. Finally, let's look take a look at inverse tangent function. Here's a graph of tangent and black, these vertical lines aren't really part of the function, they're just vertical asymptotes. So in order to get actual function, when we flip over the line y equals x, we take just one piece of the tangent function.
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Here we've taken the piece marked in black, we flip that over the line y equals x, we get this piece in red, which is actually a function because it satisfies the vertical line test. Now, you might ask, would it be possible to pick a different piece of the tangent function to invert? And the answer is yes, we could do that. And on another planet, maybe mathematicians do that. But on our planet, we use the convention that we pick this piece
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of tangent to invert, which is kind of a convenient choice since it's centered here around the origin. In the previous two examples, our choice of restricted domain for sine and for cosine was also a convention that led to a conveniently defined inverse function. In any case, based on our choice, our restricted tan function has domain from negative pi over two to pi over two. We don't include the endpoints in that interval, because our restricted tan
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function has vertical asymptotes, that negative pi over two pi over two, so it's not defined there. The range of our restricted tan function is from negative infinity to infinity. Therefore, arc tan of X has domain from negative infinity to infinity and range from negative pi over two to pi over two. Once again, tangent is taking us from angles to numbers. So arc tan is taking us from numbers to angles. For example, tangent of pi over four is one, and therefore,
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arc tan of one is pi over four. So arc tan of x means the angle between negative pi over two and pi over two whose tangent is x. y is equal to arc tan x means that x is equal to tangent of y and the Why is between negative pi over two and pi over two. The inverse tan function can also be written as 10 to the minus one of x. And once again, tan inverse of x means the inverse trig function, arc tan of x. And it's not equal to one over tan of x, which is called cotangent of x. And
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that's all for this video on the three basic inverse trig functions. sine inverse x, also known as arc sine of x, cosine inverse x, also known as arc cosine x, and tan inverse x, also known as arc tan x. In this video, we'll use implicit differentiation to find the derivatives of the inverse trig functions. First, the inverse sine function. Recall that y equals sine inverse of x means that y is the angle in radians whose sine is x. In other
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words, we can write x equals sine of y as an almost equivalent statements, I say almost equivalent, because there are lots of different y's whose side is x, lots of different angles can have the same resulting side. And for the inverse sine function, we specify that that angle y has to be between pi over two and negative pi over two, that's just the convention. Be careful not to mistake sine inverse of x, which is an inverse function,
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and one over sine x, which is a reciprocal function. These are not the same thing. This negative one does not mean reciprocal here, it means inverse function. There is another notation for inverse sine, which is arc sine. So arc sine of x is the same thing as sine inverse of x, we want to find the derivative of sine inverse of x, in other words, the derivative of y with respect to x, where y is sine inverse of x, I'm going to rewrite
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this equation here as x equals sine y, and then use implicit differentiation. So taking the derivative of both sides with respect to x, I have derivative of x is equal to the derivative of sine y. In other words, one is equal to cosine of y times dy dx. Solving for dydx, I have that dydx is one over cosine of y. Now I found the derivative, but it's not in a super useful form, because there's still a Y and expression, I'd rather have
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it all in terms of x. Well, I could rewrite this as dy dx is one over cosine of sine inverse of x, since after all, y is equal to sine inverse of x, but that's still not a super useful form, because it's difficult to evaluate this. So instead of doing this, I'm going to look at a right triangle. I want to label my triangle with y and x. Since y is my angle, I'll put it here. And since sine of y is x, and sine is opposite over hypotenuse, I can
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label my opposite side with x and my hypothesis with one. From this, I can figure out the length of my remaining side, it's going to have to be the square root of one minus x squared by the Pythagorean Theorem. Now I can compute cosine of y just from the triangle. Cosine of y is adjacent overhype hotness, so that's the square root of one minus x squared over one, or just the square root of one minus x squared. I've been implicitly assuming that
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y is a positive angle between zero and pi over two when I've been drawing this triangle, but you can check that the same formula also works. If y is a negative angle, think of it going down here on the unit circle instead of up here. So now that I have a formula for cosine y in terms of x, I can go back to my derivative and substitute and I get dy dx is one over the square root of one minus x squared. In other words, I found a formula for With the derivative of inverse sine of x, we can carry
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out a similar process to find the derivative of inverse cosine y equals cosine inverse x means that x is equal to cosine of y. And by convention, y lies between zero and pi. To find the derivative of arc cosine of x, arc cosine is just an alternative notation for cosine inverse, I can write y equals arc cosine of x, and equivalently x equals the cosine of y. And then I want to find dydx. Using implicit differentiation, please pause
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the video and try it for yourself before going on. So starting with the equation x equals cosine y, we're going to take the derivative of both sides with respect to x. The derivative of x is one, and the derivative of cosine y is negative sine y, dy dx. So d y dx is equal to negative one over sine y. As before, I can draw and label a right triangle, the angle is y. And now I know that x is cosine of y. So I'm going to put x on the adjacent
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side and one on my partner's leaving the square root of one minus x squared on the opposite side, which means that sine of y, which is opposite overhype hotness, is equal to the square root of one minus x squared. And so dy dx is going to be negative one over the square root of one minus x squared. And I have my formula for the derivative of arc cosine of x. inverse tangent can be handled very similarly. And again, you may want to
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try it for yourself before watching the video. Y equals inverse tangent of x means that x is tangent of y. And the convention is that y is supposed to lie between negative pi over two and pi over two. Proceeding as before, we write y equals our tan of x and x equals tan of y, take the derivative of both sides. So we get one equals secant squared of y d y dx. Solving for dydx we have one over seacon square root of y. And using our right triangle
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as before, we can label the angle of as y. Since I know that tangent y is x, and tangent is opposite over adjacent, I'm going to label the opposite side x and the adjacent side one, which gives us a hypothesis of the square root of one plus x squared. Now we know that secant of y is one over cosine of y. So that's going to be hi partners over adjacent. So that's the square root of one plus x squared over one. And so secant squared of y is just
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the square of this, which is one plus x squared. Now I can substitute into my formula for dy dx, and I get dy dx is one over one plus x squared, which gives me a nice formula for the derivative of inverse tangent. The other inverse trig functions cotesia inverse, seacon inverse and cosequin inverse, have derivatives that can be computed. Similarly, the following table summarizes these results. In some books, you may see absolute value signs around the
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X for the formulas for inverse secant and inverse cosecant. Of course, when x is positive, this makes no difference. And when x is negative, this discrepancy comes from differences in the convention for the range of y for these inverse trig functions seeking inverse and cosecant numbers. Notice that the derivatives of the inverse trig functions that start with CO, all have negative signs in front of them and are the negatives of the corresponding inverse trig
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functions without the CO that makes it easier to remember them. You should memorize these formulas. Let's do one example using the formulas that we just found. Let's compute The derivative of tan inverse of A plus x over a minus x, we'll want to use the formula for the derivative of tan inverse x. Now, to compute dydx, for our function, we can use the chain rule, the outside function is tan inverse, whose derivative is one over one plus the inside function squared,
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we'll need to multiply that by the derivative of the inside function. I'll just copy over the first part and take the derivative of a plus x over a minus x using the quotient rule. So I put the denominator on the bottom and square it. And then I take low times d, hi, the derivative of a plus x with respect to x is just one minus high D low, the derivative of a minus x with respect to x is negative one. I'll simplify my numerator, a minus x,
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the negative one and the negative sign here cancel, so I get plus a plus x. On the denominator, I have one plus a plus x squared over a minus x squared, multiplied by a minus x squared. canceling in the numerator, I get to a and distributing and the denominator I get a minus x squared plus a plus x squared. If I expand out the denominator, the same simplifies to two a over two A squared plus 2x squared, or just a over a squared plus x squared, which
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is a pretty nice derivative. So now you know the derivatives of the inverse trig functions. And you also know how to find them using implicit differentiation, if you ever forget them. When two or more quantities are related by an equation, then their rates of change over time are also related. That's the idea behind related rates. And this video gives an example of related rates involving distances. A tornado is 20 miles west of us, heading due east towards
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Phillips Hall at a rate of 40 miles per hour. you hop on your bike and ride due south at a speed of 12 miles per hour. How fast is the distance between you and the tornado changing after 15 minutes. In a related rates problem, it's always a good idea to draw a picture first, that can help you uncover the geometry of the problem and see how quantities are related. In this problem, we have a right triangle. Because the tornadoes traveling
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due east and the bicycles traveling due south at right angles. Let's assign variables to the quantities of interest. I'll call the distance between the tornado and Phillips Hall a. Although it starts at 20 miles, it varies with time, and therefore it's a good idea to assign it a letter a variable. I'll use B to refer to the distance between Philips Hall and the bicycle, a quantity that also varies with time. And I'll let c stand for
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the distance between the tornado and the bicycle. The problem asks us to find how fast this distance is changing. In other words, DC dt. The next step is to write down equations that relate the quantities of interest. In this problem, we know by the Pythagorean Theorem, that a squared plus b squared equals c squared. We're interested in how fast the distance between you and the tornado is changing. That's a rate of change. And the rate at which the
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bicycle is traveling and the tornado is moving. These are also rates of change. In order to work these rates of change into the problem, I'm going to take the derivative of both sides of this equation with respect to time. That's the third step. So I'm going to take DDT of a squared plus b squared. And that's equal to DDT of C squared. Notice that I'm thinking of a B and C as functions of T here, since they vary over time, on the left side, I get to a times dA DT using the chain rule,
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plus two B DB dt. And on the right side, I get to see DC dt. Now I can use the information given to me in the problem to plug in numbers and solve for the quantity of interest DC DT since the tornado was moving at a rate of 40 miles per hour The distance between the tornado and Philip's Hall is decreasing at 40 miles per hour. In other words, da dt is negative 40. That negative sign is important here, and comes from the fact that the distance
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is decreasing. Since the bicycle is moving at 12 miles per hour, the distance between Philips Hall and the bicycle is increasing at a rate of 12 miles per hour. So DBT is positive 12. The quantity is a, b and c are constantly changing. But at the time of interest, t equals 15 minutes or in hours, 0.25 hours, we can figure out what a B and C are. The tornado is starts 20 miles away, but it's moving at a rate of 40 miles per hour. So
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after a quarter of an hour, it's gone 10 miles, that means after a quarter of an hour, it's only 10 miles away. And so at the time of point two, five hours, a equals 10. The bike is moving at 12 miles per hour. So after a quarter of a mile, it's gone three miles. And so at this time, b equals three. Now using the same equation we started with, we can plug in a and b and solve for C, we know that c squared is going to be 10 squared plus three
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squared. So C is going to be the square root of 109. Plugging in the numbers into this equation, we get two times 10 times negative 40 plus two times three times 12 equals two times the square root of 109 times DC dt. So DC dt is going to be negative 800 plus 72 over two times a squared of 109, which is approximately negative 35. In other words, the distance between the tornado and us is decreasing at 35 miles per hour, the tornado is gaining on us quickly.
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These same steps will get you through a variety of related rates problems. A couple of cautionary notes. Don't plug in numbers to send. Any quantities that vary with time should be written as variables, so you can properly take the derivative with respect to time. In addition, be careful to use negative numbers for negative rates of change. That is, for quantities that are decreasing, we wouldn't have gotten the right answer if we hadn't have used a negative
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40 for the rate of change of the distance here. In this video, we solved the related rates problem, and found that riding a bicycle may not be the best way to escape a tornado. In this classic related rates problem, water's flowing into a cone shaped tank, and we have to figure out how fast the water is rising. Water flows into a tank at a rate of three cubic meters per minute, the tank is shaped like a cone with a height of four meters,
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and a radius of five meters at the top, we're supposed to find the rate at which the water level is rising in the tank. When the water height is two meters. we've drawn our picture. Now let's label some quantities of interest. It's fine to use numbers for the quantities that stay fixed throughout the problem. Like the dimensions of the tank. For any quantities that are varying with time, I need to use letters variables to represent those quantities.
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So the height of the water is varying throughout the problem. I'll call that H. And it might be handy to also talk about the radius of the part of the cone that's filled with water. I'll call that our ultimately I want to find the rate at which the water level is rising. So that's DHD T. Next, I want to write down equations that relate the quantities of interest. From geometry, I know that the volume of a cone is 1/3 times the area of the base times
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the height. So the volume of water in the cone is going to be 1/3 times pi r squared times h since h is the height of the piece of the code that contains water. And pi r squared is the area of that circular base for that piece of account. I'm calling it the base even though it's at the top. There's one more equation that's going to be handy here that comes from similar triangles. From similar triangles, we know the ratio of sides
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for the little triangle here is the same as the ratio of sides for the big triangle. In other words, we know that our over h is going to be equal to five over four. I can use this relationship to eliminate one of the variables in this equation. Let's think for a minute which one we want to eliminate. Since we're ultimately interested in finding DHD T, we need to keep the variable h in here. But since we don't have any information about how are
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you changing, it's a good idea to get rid of the R. So let's solve for r here. And we get r equals five fourths times h, and plug that back into our volume equation. So we get v equals 1/3 pi times five fourths h squared times h. Or in other words, V equals 25/48 pi h cubed. Now we're gonna derive both sides of the equation with respect to time t, to get rates of change into the problem. Remember that we're thinking of the volume of water
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and the height of water as functions of time t, we get dv dt equals 2548. It's pi times three h squared DHD t. Now let's plug in numbers and solve for the quantity of interest DHD T. From our problem, we know that water's flowing into the tank at a rate of three cubic meters per minute. So dv dt is three, we're asked to find the rate at which the water level is rising when the water height is two meters. So that's when h is two. Plugging
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in those values and solving for a DHD T, we get d h dt is equal to three divided by 2540. It's pi times three times two squared, which is 12 over 25 pi meters per second, are about point one five meters per second. This video solve the related rates problem involving volume, and use the trick of finding similar triangles to eliminate one variable. In this video, we'll do a related rates problem involving rotation and angles. a lighthouse that's half a mile west of shore, has a rotating
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light that makes two revolutions per minute in the counterclockwise direction, the shore runs north south, and there's a cave directly east of the lighthouse. How fast is the beam of light moving along the shore at a point one mile north of the cave. we've drawn a picture. Now let's label it with variables for all the quantities that are changing with time, the distance between the lighthouse and the cave that's fixed. So we don't have
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to put a variable for that. But the distance between the cave and the point on the shore where the light is hitting, that's varying. So I'll call that say x. Since we want to know how fast the beam of light is moving, we're going to want to know how that distance x is changing. In other words, we want to calculate dx dt, when x is one. The high partners of this right triangle made by the beam of light is also changing with time as as the
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angle here between the beam of light and the East West line, I'll call that angle theta. And the angle up here, I suppose is also changing and call that fee. This angle is the right angle between the East West line and the north south line. So that doesn't change, it's always 90 degrees. Next, we want to write down equations to relate the quantities of interest. Whenever I see a right triangle in a problem, I'm tempted to write down the Pythagorean theorem, which
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in this case would say one half squared plus x squared equals h squared. But in this particular problem, it doesn't look like that's going to help us much because of a tiger in theory and would relate x and h. But we don't have any information about how H is changing. The only rate of change information given to us is this two revolutions per minute. Two revolutions per minute is indirectly telling us how this angle theta is changing. Because if the light
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beam is making two revolutions per minute, then since they're two pi radians in a revolution, that amounts to a change of four pi radians per minute for the angle theta. Therefore I'd really like to write down the equation that has to do with theta and x. And from trig, I know that tangent of theta is opposite over adjacent. So I can write down tangent theta equals x divided by one half. Or in other words, tangent theta is 2x. This is
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the equation that I need that relates x and theta. Now I'm going to derive both sides with respect to time t. And I get second squared theta, d theta dt equals two times dx dt. Next, I can plug in numbers and solve for my quantity of interest, which is dx dt when x equals one. We already figured out from the two revolutions per minute, that d theta dt is four pi. Now secant theta is one over cosine theta. And since cosine theta is adjacent
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ever had partners, it's reciprocal is high partners over adjacent. So in our picture, that gives us h over one half. Well, when x equals one, H, is going to be the square root of one squared plus a half squared by the Pythagorean Theorem. And we'll divide that by one half. And simplifying, we get the square root of five fourths divided by one half, which ends up as the square root of five. So let's plug these values into our
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equation involving derivatives. And we get the square root of five squared per second squared times four pi, four d theta dt equals two times dx dt. Solving for dx dt, we get dx dt is five times four pi divided by two, or 10 pi. So what are the units here on dx dt, since our distance has been in miles, and our time is in minutes, this is 10 pi miles per minute. If I want to convert this to more standard units of miles per hour, I can just multiply my 10 pi miles per minute
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by 60 minutes per hour to get 600 pi miles per hour. That works out to about 1885 miles per hour, which is pretty darn fast. In this related rates problem, we related rotations per minute, to a change in angle per minute. And we use the trig equation to relate angle and side length. solving a right triangle means finding the length of all the sides and the measures of the angles given partial information. In this example, we're given
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the length of one side and the measure of one angle, plus we know the measure of this right angle is 90 degrees, we need to find the measure of the third angle labeled capital A, and the length of the two sides labeled lowercase b and lowercase C. To find the measure of angle A, let's use the fact that the measures of the three angles of a triangle add up to 180 degrees. So that means that 49 degrees plus 90 degrees plus a is equal to 180 degrees.
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So A is equal to 180 degrees minus 90 degrees minus 49 degrees, which works out to 41 degrees. To find the length of the side D we have a couple of possible options. We could use the fact that tan of 49 degrees, which is opposite over adjacent is B over 23. So b is 23 times tan 49 degrees, which works out to 26.46 units. Alternatively, we could use the fact that tan of 41 degrees is 23 over b since now if we're looking at the angle here 23 is our
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opposite and B is an adjacent That's a little bit harder to solve algebraically. But we can write B tan 41 degrees equals 23, which means that B is 23 divided by tan 41 degrees. With a calculator that works out again to 26.46. The reason we want to use 10 in this problem and not say sine or cosine is because 10 of say 49 degrees relates, and the unknown side that we're looking for be to the side that we know the measure of, if we had use
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sine instead, would be saying that sine of 49 is B over C, and we'd have two unknowns, which would make it difficult to solve. Next, to find the side length C, we can have a few options, we could use a trig function again, for example, we could use the cosine of 49 degrees, that's adjacent over hypotenuse, which is 23 oversee. Solving for C, we get that C is 23 over cosine 49, which works out to 35.06 units. Another option would be to
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use the Pythagorean Theorem to find C. Since we know 23 squared plus b squared equals c squared. In other words, that's 23 squared plus 26.46 squared equals c squared, which means that C is the square root of that song, which works out again, to 35.06. To review, the ideas we used were, the sum of the angles is equal to 180 degrees. We used facts like tangent of an angle being opposite over adjacent and similar facts about sine and cosine. And
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we use the Pythagorean Theorem. This allowed us to find all the angles and side lengths of the triangle, knowing just the side length of one side and the angle of one of the non right angles to begin with. In this next example, we don't know any of the angles except for the right angle, but we know to have the side lengths. To find the unknown angle theta, we can use the fact that cosine theta is adjacent overhype
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hotness, so that's 10 over 15. Cosine is a good trig function to use here, because this equation relates our unknown angle to our two known sides. So we just have one unknown in our equation to solve for. To solve for theta, we just take the cosine inverse of 10/15, which is 0.8411 radians, or 48.19 degrees. To find the measure of angle fee, we could use the fact that sine of fee is 10 over 15 and take sine inverse of 10/15. But probably
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a little easier, let's just use the fact that these three angles do 180 degrees. That tells us that fee plus 90 plus 48.19 is equal to 180. Which means that fee is 41.81. Finally, we can find x either using a trig function, or by using the Pythagorean Theorem. To find it using a trig function, we could write down something like tan of 48.19 degrees is x over 10. To find that using the Tyrion theorem, we'd write down 10 squared plus x squared
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equals 15 squared. I'll use a Pythagorean theorem and find the x by doing the square root of 15 squared minus 10 squared. That gives me an answer of 11.18. Notice that we use many of the same ideas as in the previous problem. For example, the fact that the sum of the angles is 180. The Pythagorean theorem and the trig functions like tan, sine and cosine, we also use the inverse trig functions to get from an equation like this one to the
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angle. This video showed how it's possible to find the length of all the sides of a right triangle, and the measures of all the angles given partial information. For example, the measure of one angle and one side or from two sides. This video gives some definitions and facts related to maximum and minimum values. functions. function f of x has an absolute maximum at the x value of C. If f of c is greater than or equal to f of x for all x
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in the domain of f. The point with x&y coordinates of C FRC is called an absolute maximum point. And the y value f of c is called the absolute maximum value. Now, if I draw a graph of f, the y value f of c is the highest value that that function ever achieves. And an absolute maximum point is just a point where it achieves that maximum value. Now, it's possible for a function to have more than one absolute maximum point, if there happens to be a tie
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for the highest value. But a function has at most one absolute maximum value. A function f of x has an absolute minimum that x equals C, if f of c is less than or equal to f of x, for all x in the domain of f. In this case, the point c f of c is called an absolute minimum point. And the y value f of c is called the absolute minimum value. In the graph of f of x, f of c is now the lowest point that the function achieves anywhere on its domain,
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and C SOC, are the coordinates of a point where the function achieves that minimum value. For example, this function has an absolute minimum value of about negative eight has an absolute minimum point with coordinates three, negative eight. If this function stops here, and just has a domain from zero to four, then the function has an absolute maximum value of 10 at the absolute maximum point with coordinates for 10. If however, the function
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keeps going in this direction, it will not have an absolute maximum value at all. absolute maximum and minimum values can also be called global maximum and minimum values. In addition to absolute maximum mins, we can talk about local maximums. So a function f of x has a local maximum at x equals C. If f of c is greater than or equal to f of x, for all x, near C. By near C, we mean there's some open interval around C for which this
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was true. For our graph of f, we have a local maximum right here. Even though it's not the highest point anywhere around since there's a higher point up here, this is the highest point in an open interval around see the point C FFC is called a local maximum point. And the y value f FC is called a local maximum value. Similarly, a function f of x has a local minimum at x equals C, if f of c is less than or equal to f of x for all x, near
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C. And the point c f of c is called a local minimum point. And the y value f of c is called a local minimum value. A function might have many local minimum values. In this example, assuming that the domain is zero to four, we have a local minimum point right here. Because it's the lowest point anywhere nearby. It also happens to be an absolute minimum point. Now turning our attention to local maximums, we have a local maximum point right
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here with coordinates about one two. Since f of one is as high or higher than f of x for any x value in an open interval around one. In this example, the absolute maximum point of 410 does not count as a local maximum point. Simply because we can't take an open interval on both sides of for the function doesn't exist on the right side. And so for that sort of technical reason, we end up with an absolute maximum point That's not a local
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maximum point here. local maximum and minimum values can also be called relative maximum and minimum values. Please take a look at this graph and pause the video for a moment to mark all local maximum minimum points, as well as all global, that is absolute maximum points. See if you can find the absolute maximum value and the absolute minimum value for the function. I'm going to mark the local maximum points in green, and the absolute maximum
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points in red. The function definitely has a local men here. Since this is the lowest point anywhere nearby and then open interval, and there's a local max point here. There's also a local min point here, where the function also hits a low point and open interval. But that local man is also an absolute man. So I'll mark it half green and half red. There's also a local min point here at the point three, two,
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since this point is the as low or lower than any point in an open interval. And the function is defined in an open interval around three, even though it's discontinuous there. In fact, this point is tied for local minimum, with all the points on this interval here, between two and three, there are as low or lower than all points in an open interval around them. The point 04 doesn't count as a local max, because the function is not defined on the
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other side of zero. So there's no open interval to to consider. This point is also not an absolute maximum because the function gets higher over here. In fact, as long as this trend continues, the function f of x has no absolute maximum value at all, because its values just keep getting higher and higher as x goes off to infinity. There's one more point that I want to consider. And that's this point here, at three, three and a half.
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Well, it's tempting to say that f has a local maximum here, it looks like it's the highest point in the ground. But in fact, there is no point here at three 3.5. Right, the functions value at three is actually down here, too. So there's no point here to be a local maximum point. And if you start looking at points really close to that point, those aren't local maximums either, because you can always find a point just a little bit higher as you get
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closer and closer, but don't quite reach this missing point of three 3.5. So we have all the absolute and local maximum points marked. And now to find the absolute maximum value. Well, we just said that there is none. But the absolute minimum value is the y value of this absolute minimum point here. So I'd say that's about 0.5. Here I've drawn the graph of a function. What do you notice about the derivative of this function at its local
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maximum and minimum points, please pause the video and think about it. Well, the local maximum minimum points are here, here and here. And at two of those points, the derivative f prime of c equals zero. And that the third point, f prime of c does not exist, because the function has a corner. A number c is called a critical number for a function f if f prime of c does not exist, or f prime of c exists and equals zero. So in other words, all of
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these local maximum minimum points for this example, they're all critical points. And this is true in general, if f has a local max or min at C, then C must be a critical number for F. We also say that the point c f of c is a critical point for F. It's important not to read too much into this statement. The statement says that if f has a local max or man at C then C must be a critical number. But the converse doesn't hold. In other words,
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if c is a critical number, then f may or may not have a local max or men at sea. One example to keep in mind is the function f of x equals x cubed at a value of C of zero. Since f prime of x is 3x squared, we have that f prime of zero equals zero. So zero is a critical number. But notice that F does not have a local maximum man at x equals zero. In this video, we defined absolute and local, maximum and minimums. We also defined critical
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numbers, which are numbers c, where f prime of c equals zero, or f prime of c does not exist. We noted that if f has a local max or min at C, then C is a critical number. But not necessarily advice. In this video, we'll see how the first derivative and the second derivative can help us find local maximums and local minimums for a function. Recall that f of x has a local maximum at x equals C. If f of c is greater than or equal to f
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of x, for all x, in an open interval around C, F of X has a local minimum at x equals C. If f of c is less than or equal to f of x, for all x, in an open interval around C. In this example, the function f has a local maximum at x equals six at x equals 11. And a local minimum at x equals about 10. We've seen before that if f has a local max or local min at x equals C, then f prime of c is equal to zero or it does not exist. Number C at
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which f prime of c is zero or does not exist, are called critical numbers. But you have to be careful, because it is possible for F to have a critical number at C. That is a place where if privacy is equal to zero, or does not exist, but not have a local max, or min at x equals C. In fact, this happens in the graph above, at x equals two, since f prime of two is zero, but there's no local max or min there. Please pause the video for
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a moment and try to figure out what's different about the derivative of f in the vicinity of x equals two, where there's no local max or min, and in the vicinity of x equals 610 and 11 where there are local maxes and mins. Near the critical point at x equals two, the derivative is positive on the left and positive again on the right. But near the local maximums, the derivative is positive on the left and negative on the right. And near the local
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minimum, the derivative is negative on the left and positive on the right. These observations help motivate the first derivative test for finding local maximums and minimums. The first derivative test says that if f is a continuous function near x equals C, and if c is a critical number, then we can decide if f has a local maximum or minimum at x equals c by looking at the first derivative near x equals C. More specifically, if we know that f prime of x
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is positive for x less than c, and negative for x greater than c, then our function looks something like this. Or maybe like this, your x equals C, and so we have a local max at x equals C. If on the other hand, f prime of x is negative for x less than c, and positive for x greater than c, then our function looks something like this. Or maybe like this, your x equals C and so we have local men at x equals see. If our first derivative is positive on
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both sides of C or negative on both sides of C, then we do not have a look All extreme point at all at x equals C. Instead, our graph might look something like this, or maybe like this. The first derivative test is great, because it lets us locate local extreme points just by looking at the first derivative. The second derivative test gives us an alternative for finding local maximum points by using the second derivative. Specifically, the second
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derivative test tells us that if f is continuous near x equals C, then if f prime of c is equal to zero, and f double prime of c is greater than zero, then f has a local min at x equals C. If on the other hand, f prime of c equals zero, and f double prime of c is less than zero, then f has a local max at x equals C. Note that if f double prime of c is equal to zero, or does not exist, then the second derivative
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test is inconclusive. We might have a local max or a local man at x equals C, or we might not. So we'd have to use a different method like the first derivative test to find out. In this video, we introduced the first derivative test and the second derivative test, which allow us to determine if a function has a local minimum or a local maximum at a certain value of x. In this video, I'll work through two examples of finding extreme values, that
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is, maximum values and minimum values of functions. In the first example, we're asked to find the absolute maximum and minimum values for this rational function g of x on the interval from zero to four, these maximum and minimum values could occur at critical numbers in the interior of the interval, or they could occur at the endpoints of the interval. So we'll need to check the critical numbers, and check the endpoints and compare our values.
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To find the critical numbers, those are the numbers where g prime of x is equal to zero or does not exist. So let's take the derivative g prime of x using the quotient rule. So we get x squared plus x plus two squared on the denominator, and then we have low times d high, the derivative of the numerator is one minus high times the derivative of the denominator, that's 2x plus one. Before we figure out where that zero or doesn't exist, let's simplify
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it a little bit. So we can multiply out the numerator. I'll distribute the negative sign. And I'll add together like terms in the numerator, I'm just leaving the denominator alone on all these steps. So our simplified numerator is going to be minus x squared plus 2x plus three. Now that I've simplified the derivative, I can figure out where it's equal to zero and where it doesn't exist. Let me clear a little space. Now the only way that g prime
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of x could not exist, is if the denominator is zero. But on our interval, where x is between zero and 4x squared plus x plus two is always greater than or equal to two. So the denominator is never zero on this interval. In fact, it turns out that x squared plus x plus two is never zero, even if we look outside this interval. And you can check that if you want to use the quadratic formula. But in any case, we don't have to worry about the places where
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g prime of x does not exist. So we only have to worry where g prime of x is equal to zero. To find where g prime of x is equal to zero, we just have to check where the numerator is equal to zero. So I'll set negative x squared plus 2x plus three equal to zero and multiply both sides there by negative one and a factor I get that x equals three or x equals minus one. So these are my critical numbers. But notice that one of these critical numbers
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negative one doesn't even lie within my interval, so I don't have to worry about it. All I have to worry about is x equals 3x equals three is one place where my function g could have an absolute maximum or minimum. So let's figure out G's value there by plugging in three for x that evaluates to to 14th, or one set So we've checked the critical numbers. Now let's go ahead and check the endpoints. Those are the point the x values of zero, and four,
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since our interval is from zero to four. Plugging in, we get that g of zero is negative one half, and g of four is 320 seconds. I sometimes like to make a table of all these candidate values. The candidate x values are 03, and four and the corresponding g of x values we found were negative a half 1/7, and 320 seconds. Now to find the absolute maximum and minimum values, all I have to do is figure out which one of these y values is the biggest and which is
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the smallest. Well, clearly negative one half is a smallest, so that's the absolute minimum value. And we just need to compare 1/7 and 320 seconds to see which is bigger. Now, 1/7 is the same as 320 firsts, which is going to be bigger than 320 seconds. So one half 1/7 is our absolute max value. We can confirm this by looking at a graph of our function g. Remember, we're just interested in the interval from zero to four. So we're just
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interested in this section of the graph. And it does look like the minimum value is here at when x equals zero minimum value of negative one half like we found, and the maximum value. Well, I'm not sure exactly where it is from this graph, but it does look like it's somewhere around three. And that is a value of something around 1/7. So the graph does confirm what we found as a more precise answer using calculus. For the next example, let's find the absolute
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extreme values for the function f of x, which is the absolute value of x minus x squared, on the interval from negative two to two. As before, we can find absolute extreme values by checking first the critical numbers, and then also the endpoints of the interval negative two and two. To find the critical numbers, we need to take the derivative of our function. But because our function involves the absolute value, it's a little tricky to take the derivative.
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Instead, let's first rewrite f using piecewise notation. Recall that, if we're looking at the absolute value of x, when x is bigger than or equal to zero, absolute value of x is just x. So f of x will be x minus x squared. On the other hand, when x is less than zero, the absolute value of x is negative x. So f of x will be negative x minus x squared. Now to take the derivative, we can take the derivative of each piece. So when x is bigger
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than zero, I don't want to take the derivative when x equals zero, because there might be funny things happening, you know, a cost per corner, so I'm just gonna worry about when x is bigger than zero, and when x is less than zero, for now, when x is bigger than zero, I can just use the power rule I get one minus 2x is the derivative when x is less than zero, I get negative one minus 2x. And now to find where I have critical numbers,
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I need to find where f prime of x is equal to zero, or f prime of x does not exist. Well, f prime of x equals zero, where one minus 2x equals zero for x bigger than zero. And where one, negative one minus 2x is equal to zero for x less than zero. So that corresponds to x equal one half for x bigger than zero, and x equals negative one half for x less than zero. So those are my first two critical numbers. And that is they do lie within my
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interval that I'm interested in. But I also have to worry about where f prime of x does not exist. And the candidate x value for that is where x equals 01 way to convince ourselves the derivative does not exist when x equals zero is to look at the fact that the derivative is very close to one for x values, very close to zero from the right side and very close to negative one for x values from the left side. So the graph of the function is going
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to have to be sloping down with a slope near negative one for x less than zero and up with a slope near one for x greater than zero, and so it'll end up having a cost per corner there. Also notice that even if I weren't 100% sure that the derivative didn't exist at x equals zero, it's not going to hurt to consider this x equals zero as a possible additional candidate for the absolute max or min value. So I've got my three critical numbers, and my end
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points are just going to be x equals negative two and x equals two. So let me make a chart of values my x values to consider are negative two, negative one half 01, half and positive two, and my corresponding f of x values are going to be, let's say absolute value of negative two minus negative two squared works out to two minus four, which is negative two, I plug in negative one half, I get one half minus 1/4, which is 1/4. plug in zero, I get zero.
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and plugging in one half, I get 1/4. And plugging in two, I get negative two again. So now my biggest value is going to be 1/4. So that's my absolute max value, and my smallest value is going to be negative two. So that's my absolute min value, I can confirm what I found looking at the graph. So here I've graphed my function, y equals absolute value of x minus x squared on the interval from negative two to two. And I can see indeed, that my
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absolute min is going to be a value of negative two, it occurs at two absolute minimum points, and my absolute maximum is going to be a value of about 1/4. And that occurs at two absolute maximum points. And that concludes this video on finding extreme values. the mean value theorem relates the average rate of change of a function on an interval to its instantaneous rate of change, or derivative. Let's assume that f is a function defined on a closed interval
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a, b, and maybe defined in some other places to let's assume that f is continuous on the whole closed interval. And that is differentiable on the interior of the interval, then the mean value theorem says that there must be some number c in the interval a, b, such that the average rate of change of f on a B is equal to the derivative of f at C. In symbols, we can write the average rate of change as f of b minus F of A over B minus A. And that
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has to equal f prime at C for some number C. On the graph, the average rate of change of f is the slope of the secant line. And so the mean value theorem says that there's some number c, somewhere in between a and b, so that the slope of the secant line is exactly the same as the slope of the tangent line at that x value of C. The number c is not necessarily unique. So I encourage you to pause the video and see if you can draw
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a graph of a function where there's more than one c value that works. So you might have drawn something maybe like this. Now, if we draw our secant line, there's two values of c, where the slope of the tangent line is equal to the slope of the secant line. In this example, we're asked to verify the mean value theorem for a particular function on a particular interval. Verify means that we need to check the hypotheses of the theorem
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hold. And also the the conclusion holds. The hypotheses are that f is continuous on the closed interval, one three, and that is differentiable on that interior of that interval. Both of these facts are true, because f is a polynomial. Now we need to verify that the conclusion of the mean value theorem holds. In other words, we need to find a number c in the interval one, three, such that the derivative of f at C is equal to the average rate of change of f on the interval from one to three.
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Now f prime of x is 6x squared minus eight. So f prime at any number c is just six c squared minus eight. We can also compute f of three just by plugging in and get 31 and f of one is negative five. Plugging in these values into our equation, we get that six c squared minus eight has to equal 31 minus negative five over To, in other words, six c squared minus eight had better equal 18, which means that six c squared needs to equal 24. So C
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squared has to equal four, which means that C has to equal plus or minus two. Since negative two is not in the interval from one to three, we're left with a c value of positive two. So C equals two is the number we're looking for. And at C equals to f prime is equal to 18, which is the average rate of change of f on the interval. we've verified the mean value theorem. In this example, we're told that f of one is seven, and that the derivative
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of f is bounded between negative three and negative two. On the interval one six, we're asked to find the biggest and smallest values that f of six could possibly be. Well, the mean value theorem gives us one way of relating the derivative of the function to its values on the endpoints of the interval. More specifically, the mean value theorem tells us that the average rate of change F of six minus f of one over six minus one is equal to the derivative f
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prime of c, for some C, in the interval, one, six. Since the derivative is bounded between negative three and negative two, we know that the average rate of change is bounded between negative three and negative two. We know that f of one is seven. And now we can solve this inequality for f of six. Multiply the inequality by five and add seven. And now we can see that negative eight is the smallest possible value for F six and negative three is the
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largest possible roles, there is an important special case of the mean value theorem. If f is a function defined on the closed interval a b, and f of x is continuous on that whole closed interval, differentiable on the interior of the interval. And if f of a is equal to f of b, then there's a number c in the interval a, b, such that f prime of c is zero. If we look at a graph of such a function that has equal values at a, and at B, we can see where its
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derivative has to be zero at a maximum, or a minimum in between A and B. To see why the rolls there is a special case of the mean value theorem. Think about what the mean value theorem would say about this function, it would say there is a C, such that f prime of c is equal to the average rate of change of the function. But since f of b and F of A are the same, by our assumption, this average rate of change is just zero. And so the mean
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value theorem, its conclusion is that there's a C, such that f prime of c equals zero, which is exactly the conclusion of rules theorem. In this video, we saw that for a function that's continuous on a closed interval, and differentiable on the interior of that interval, the average rate of change of the function is equal to the instantaneous rate of change of the function, f prime of c for some C in the interval. This video gives two proofs
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of the mean value theorem for integrals. the mean value theorem for integrals says that for continuous function f of x, defined on interval from a to b, there's some number c between A and B, such that f of c is equal to the average value of f. The first proof that I'm going to give uses the intermediate value theorem. Recall that the intermediate value theorem says that if we have a continuous function f defined on an interval, which I'll
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call x 1x, two, if we have some number l in between, f of x one and f of x two, then f has to achieve the value l somewhere between x one and x two. Keeping in mind the intermediate value theorem, let's turn our attention back to the mean value theorem for integrals. Now, it's possible that our function f of x might be constant on the interval from a to b. But if that's true, then our mean value theorem for integrals holds easily, because f AV is
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just equal to that constant, which is equal to f of c for any c between A and B. So let's assume that f is not constant, will it continue continuous function on a closed interval has to have a minimum value and a maximum value, which I'll call little m, and big M. Now, we know that F's average value on the interval has to be between its maximum value and its minimum value. If you don't believe this, consider the fact that all of F's values on
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the interval have to lie between big M and little m. And if we integrate this inequality, we get little m times b minus a is less than or equal to the integral of f is less than or equal to big M times b minus a. Notice that the first and the last integrals, were just integrating a constant. Now if I divide all three sides by b minus a, I can see that little m is less than or equal to the average value of f is less than or equal to big M
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as I wanted. Now, I just need to apply the intermediate value theorem with F's average as my number L and little m and big M as my values of f of x one and f of x two. The intermediate value theorem says that F average is achieved by f of c for some C in between my x one and x two. And therefore, for some C in my interval a b. And that proves the mean value theorem for integrals. Now I'm going to give a second proof for the mean value theorem for integrals.
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And this time, it's going to be as a corollary to the regular mean value theorem for functions. Recall that the mean value theorem for functions, says that if g of x is continuous on a closed interval, and differentiable on the interior of that interval, then there's some number c in the interval, such that the derivative of g at C is equal to the average rate of change of G, across the whole interval from a to b. Let's keep the mean value theorem for functions
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in mind, and turn our attention back to the mean value theorem for integrals. I'm going to define a function g of x to be the integral from a to x of f of t dt, where F is the function given to us in the statement of the mean value theorem for integrals. Notice that g of A is just the integral from a to a, which is zero, while g of B is the integral from a to b of our function. Now, by the fundamental theorem of calculus, our function g of x is
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continuous and differentiable on the interval a, b, and g prime of x is equal to f of x. And by the mean value theorem for functions, we know that g prime of c has to equal g of b minus g of a over b minus a, for some numbers, C and the interval a b, if we substitute in the three facts above, into our equation below, we get f of c is equal to the integral from a to b of f of t dt minus zero over b minus a, which is exactly the conclusion that we
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wanted to reach. This shows that the mean value theorem for integrals really is the mean value theorem for functions where our function is an integral. And this completes the second proof of the main value theorem for integrals. So now I've proved the mean value theorem for integrals in two different ways. And I've used a lot of the great theorems of calculus along the way in this video, We'll solve inequalities involving polynomials like
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this one, and inequalities involving rational expressions like this one. Let's start with a simple example, maybe a deceptively simple example, if you see the inequality, x squared is less than four, you might be very tempted to take the square root of both sides and get something like x is less than two as your answer. But in fact, that doesn't work. To see why it's not correct, consider the x value of negative 10. Negative 10 satisfies the
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inequality, x is less than two, since negative 10 is less than two. But it doesn't satisfy the inequality x squared is less than four, since negative 10 squared is 100, which is not less than four. So these two inequalities are not the same. And it doesn't work to solve a quadratic inequality just to take the square root of both sides, you might be thinking part of why this reasoning is wrong, as we've ignored the negative two option, right? If
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we had the equation, x squared equals four, then x equals two would just be one option, x equals negative two would be another solution. So somehow, our solution to this inequality should take this into account. In fact, a good way to solve an inequality involving x squares or higher power terms, is to solve the associated equation first. But before we even do that, I like to pull everything over to one side, so that my inequality has
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zero on the other side. So for our equation, I'll subtract four from both sides to get x squared minus four is less than zero. Now, I'm going to actually solve the associated equation, x squared minus four is equal to zero, I can do this by factoring 2x minus two times x plus two is equal to zero. And I'll set my factors equal to zero, and I get x equals two and x equals minus two. Now, I'm going to plot the solutions to my equation
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on the number line. So I write down negative two and two, those are the places where my expression x squared minus four is equal to zero. Since I want to find where x squared minus four is less than zero, I want to know whether this expression x squared minus four is positive or negative, a good way to find that out is to plug in test values. So first, I plug in a test value in this area, the number line, something less than negative two, say x equals negative three. If I plug in negative three
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into x squared minus four, I get negative three squared minus four, which is nine minus four, which is five, that's a positive number. So at negative three, the expression, x squared minus four is positive. And in fact, everywhere on this region of the number line, my expression is going to be positive, because it can jump from positive to negative, without going through a place where it's zero, I can figure out whether x squared minus four is positive or
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negative on this region, and on this region of the number line by plugging in test value similar way, evaluate the plug in between negative two and two, a nice value is x equals 00 squared minus four, that's negative four and negative number. So I know that my expression x squared minus four is negative on this whole interval. Finally, I can plug in something like x equals 10, something bigger than two, and I get 10 squared minus four. Without even
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computing that I can tell that that's going to be a positive number. And that's all that's important. Again, since I want x squared minus four to be less than zero, I'm looking for the places on this number line where I'm getting negatives. So I will share that in on my number line. It's in here, not including the endpoints, because the endpoints are where my expression x squared minus four is equal to zero and I want it strictly less than zero, I can write
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my answer. As an inequality, negative two is less than x is less than two, or an interval notation as soft bracket negative two, two soft bracket. Our next example, we can solve similarly, first, we'll move everything to one side so that our inequality is x cubed minus 5x squared minus 6x is greater than or equal to zero. Next, we'll solve the associated equation by factoring. So first, I'll write down the equation. Now I'll factor out an
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x. And now I'll factor the quadratic. So the solutions to my equation are x equals 0x equals six and x e equals negative one, I'll write the solutions to the equation on the number line. So that's negative one, zero, and six. That's where my expression x times x minus six times x plus one is equal to zero. But I want to find where it's greater than or equal to zero. So again, I can use test values, I can plug in, for example, x equals negative
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two, either to this version of expression, or to this factored version. Since I only care whether my answer is positive or negative, it's sometimes easier to use the factored version. For example, when x is negative two, this factor is negative. But this factor, x minus six is also negative when I plug in negative two for x. Finally, x plus one, when I plug in negative two for x, that's negative one, that's also negative. And a negative
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times a negative times a negative gives me a negative number. If I plug in something between negative one and zero, say x equals negative one half, then I'm going to get a negative for this factor, a negative for this factor, but a positive for this third factor. Negative times negative times positive gives me a positive for a test value between zero and six, let's try x equals one. Now I'll get a positive for this factor a negative
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for this factor, and a positive for this factor. positive times a negative times a positive gives me a negative. Finally, for a test value bigger than six, we could use a x equals 100, that's going to give me positive positive positive. So my product will be positive. Since I want values where my expression is greater than or equal to zero, I want the places where n equals zero. And the places where it's positive. So my final answer will
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be close bracket negative one to zero, close bracket union, close bracket six to infinity. As our final example, let's consider the rational inequality, x squared plus 6x plus nine divided by x minus one is less than or equal to zero. Although it might be tempting to clear the denominator and multiply both sides by x minus one, it's dangerous to do that, because x minus one could be a positive number. But it could also be a negative number. And when
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you multiply both sides by a negative number, you have to reverse the inequality. Although it's possible to solve the inequality this way, by thinking of cases where x minus one is less than zero or bigger than zero, I think it's much easier just to solve the same way as we did before. So we'll start by rewriting so that we move all terms to the left and have zero on the right, well, that's already true. So the next step would be to solve the
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associated equation. That is x squared plus 6x plus nine over x minus one is equal to zero. That would be where the numerators 0x squared plus 6x plus nine is equal to zero, so we're x plus three squared is zero, or x equals negative three, there's one extra step we have to do for rational expressions. And that's we need to find where the expression does not exist. That is, let's find where the denominator is zero. And that said, x
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equals one. I'll put all those numbers on the number line, the places where my rational expression is equal to zero, and the place where my rational expression doesn't exist, then I can start in with test values. For example, x equals minus four, zero and two work. If I plug those values into this expression here, I get a negative answer, a negative answer and a positive answer. The reason I need to conclude the values on my number line
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where my denominator is zero is because I can my expression can switch from negative to positive by passing through a place where my rational expression doesn't exist, as well as passing by passing flew to a place where my rational expression is equal to zero. Now I'm looking for where my original expression was less than or equal to zero. So that means I want the places on the number line where my expression is equal to zero, and also the
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places where it's negative. So My final answer is x is less than one, or an interval notation, negative infinity to one. In this video, we solved polynomial and rational inequalities by making a number line. And using test values to make a sign chart. The first and second derivative of a function can tell us a lot about the shape of the graph of the function. In this video, we'll see what f prime and f double prime can tell us about where the
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function is increasing and decreasing, is concave up and concave down and has inflection points. We say that a function is increasing. If f of x one is less than f of x two, whenever x one is less than x two. In other words, the graph of the function goes up. As x increases from left to right, we say the function f is decreasing. If f of x one is greater than f of f two, whenever x one is less than x two. In other words, the height of the function
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goes down as we move from left to right. In this graph, it's a little hard to say what's happening when x is near two, is it completely horizontal, or is the graph slightly increasing? If we assume it's slightly increasing, then, in this example, f of x is increasing as x ranges from zero to six, and again as x ranges from 10 to 11. The graph is decreasing for x values between six and 10. And for x values between 11 and 12, the first derivative of
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f can tell us where the function is increasing and decreasing. In particular, if f prime of x is greater than zero for all x on an interval, then f is increasing on this interval. This makes sense, because f prime being greater than zero means the tangent line has positive slope. Similarly, if f prime of x is less than zero for all excellent interval, then f is decreasing on this interval. That's because a negative derivative means the tangent line has a negative slope.
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A precise proof of these facts can be found in the textbook, or in another video, we say that a function is concave up on an interval from a to b. If informally, it looks like a bowl that could hold water on that interval. More formally, the function is concave up on that interval. If all the tangent lines for the function on that interval, lie below the graph of the function. The function is concave down on the interval from a to b.
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If informally, it looks like an upside down ball that would spill water on that interval. Or more formally, the function is concave down. If all the tangent lines lie above the graph of the function on an interval. In this example, f is concave up around here. And again around here. On the left piece, it looks like part of a ball that could hold water. So we can say that f is concave up on the intervals from two to four, and the interval
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from eight to 11. f is concave down on this piece, and this piece and this piece, so we can say that f is concave down on the interval from zero to two, from four to eight, and from 11 to 12. The concavity of a function is related to its second derivative. Here, where the function is concave up, its derivative is going from essentially zero to larger positive values. So the first road was increasing, which means the second derivative is positive.
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On this section of the graph, which is also concave up, the driven is going from negative values to zero. That's an increase in the first derivative. So that means the second derivative here must be positive. And in this piece, where the first term is going from zero to positive values, the first derivative is also increasing, so the second derivative is also positive. On the parts of the function that are concave down, we can see that in
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this example, the second derivative is negative. Here, the first derivative is going from positive towards zero, that's a decrease in the first derivative or a negative second derivative. Here the first derivative is going from positive to zero to negative, that's also a decreasing first derivative or a negative second derivative. And the same thing happens on this section here. In general, we can use the second derivative to predict the concavity of a function. The
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concavity test says that if the second derivative is positive for all x on an interval, then the function f is concave up on that interval. Similarly, if the second derivative is negative for all x on the interval, then the function f is concave down on that interval. One way to remember the concavity test is that a positive second derivative gives us a happy face. So the smile is supposed to be a concave up function. And a negative second derivative gives us
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a sad face where the smile or the frown, I guess, is a concave down function. Next, let's talk about inflection points. A function has an inflection point at x equals C, if it's continuous at C, and it changes concavity at C. In other words, f has an inflection point at x equals C. If f changes from concave up to concave down at x equals C, or it changes from concave down to concave up. In this graph, if we draw the concavity regions again, we
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see that F has an inflection point at x equals two, where the function changes from concave down to concave up at x equals four, where the function changes from concave up to concave down at x equals eight, and again, at x equals 11. Since concavity, has to do with the second derivative being positive or negative, inflection points happen where the second derivative changes sign from positive to negative, or from negative to positive. And that's exactly
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what the inflection point test says. If f double prime of x changes sign at x equals C, then f has an inflection point at x equals C. Now in order to change from positive to negative or negative positive, f double prime has to go through zero, or go through a point where it doesn't exist. But you have to be careful, just because f double prime is zero, it doesn't exist, does not guarantee that you necessarily have an
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inflection point, because it could be zero and still be positive on both sides or negative on both sides. For example, if f of x is x to the fourth, then f prime of x is for x cubed, and f double prime of x is 12x squared. So f double prime at zero is certainly zero. But there is no inflection point at x equals zero. In fact, the graph of f of x equals x to the fourth looks kinda like a flattened quadratic, and so there's no change in calm
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cavity, f is concave up on both sides of x equals zero. In this video, without the first derivative can tell us where the function is increasing and decreasing, while the second derivative can tell us where the function is concave up and concave down. And the second derivative, changing sign from positive to negative or negative or positive can tell us where we have inflection points. Since lines are much easier to work with, and more
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complicated functions, it can be extremely useful to approximate a function near a particular value with its tangent line. That's the central idea of this video. Let's start with an example. Suppose that F of T is the temperature in degrees Fahrenheit at time t measured in hours, where t equals zero represents midnight. Suppose that f of six is 60 degrees, and the derivative f prime of six is three degrees per hour. What's your best estimate for the temperature
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at 7am and at 8am? Please pause the video for a moment to make your estimate. The temperature at 6am is 60 degrees. So the temperature at 7am, which we're calling F of seven is approximately 60 degrees. But we can do better than this. At 6am, the temperature is rising, in fact, it's rising at a rate of three degrees per hour. If this rate of change continues, then by 7am, the temperature will have risen three degrees and reached 63 degrees. And by 8am,
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the temperature will have had two hours to rise from 60 degrees by a rate at a rate of three degrees per hour. So f of eight should be about 60 degrees plus the three degrees per hour times two hours or 66 degrees. These estimates use all the information, we're given both the value of the temperature at six, and its rate of change. Let's see what these estimates mean graphically, in terms of the tangent line. I'll draw a rough graph of temperature
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over time. And I'll also draw in the tangent line at time six. At time six, the height of the function and the tangent line is equal to 60 degrees. The tangent line has slope three degrees per hour. So that's a rise of a run of three, which means that seven o'clock, which is one hour, after six o'clock, the tangent line has risen by three degrees. And at eight o'clock, the tangent line has risen by another three degrees. So at seven o'clock,
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our tangent line has had 63 degrees, and at eight o'clock, our tangent line has height 66 degrees. When making these estimates, here, we were actually using the tangent line to approximate our function. Our actual temperature function may be rising more steeply than the tangent line, or it possibly could be rising less steeply, like in this picture. But either way, the tangent line is a good approximation for our function. When time is near six o'clock.
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The idea of approximating a function with its tangent line is a very important idea that works for any differentiable function. Let f of x be any differentiable function and let A be an arbitrary x value. Let's suppose we know the value of f at a, we'll call it F of A. And let's say we want to find the value of f, add an x value near a, let's call it a plus delta x where delta x means a small number. If we can't compute f of a plus delta
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x directly, we can try to approximate it using the tangent line. We know that the tangent line has a slope given by f prime of a. And so when we go over by a run of delta x, the tangent line goes up by a rise of f prime of A times delta x. So the height of the tangent line is going to be f of a plus f prime of A times delta x. The linear approximation principle says that we can approximate our function with
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our tangent line. In other words, f of a plus delta x is approximately equal to f of a plus f prime of a delta x. Remember that delta x is supposed to be a small number, because if you get too far away from a, your tangent lines no longer going to be a good approximation of your function. But how small is small enough is sort of a judgment call. Sometimes the approximation principle is written with different symbols, if we let x equal a plus delta x,
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so x is a number close to a, then delta x is x minus a. And we can rewrite the approximation principle, as f of x is approximately f of a plus f prime of A times x minus a. The quantity on the right side here is sometimes referred to as l of x, and called the linearization of f at A. That is the linearization of f of a is l of x, which is equal to f of a plus f prime of A times x minus a. So the approximation principle can also be written as f of x is
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approximately equal to l of x. Let's look a little more closely at this linearization equation and what it has to do with the tangent line. Suppose we were going to try to write down the equation of the tangent line at x equals a, well, the equation for any line can be given in point slope form as y minus y naught equals the slope times x minus x naught. Since we're looking for the tangent line that goes through the point A f of a,
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we can set x naught equal to a and y naught equal to f of a. Also, the slope of the tangent line is just f prime of a. So we can rewrite this as y minus F of A equals f prime of A times x minus a solving for y, we get y equals f of a plus f prime of A times x minus a. So this equation for the tangent line is really just the equation that we have for the linearization. But linearization is really just a fancy word for the tangent line. There's a lot of notation
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and definitions on this page. But there's only one important principle that you need to remember. And that's the idea that you can approximate a function with its tangent line. If you can keep that idea and this picture in mind, then it's easy to come up with this approximation principle. And its alternative forms. Let's use the approximation principle in an example, the approximation principle tells us that f of a plus delta x is approximately
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equal to f of a plus f prime of A times delta x, we need to figure out what f should be what a should be, and what delta x should be. Since we're trying to figure out the square root of 59, it makes sense to make our function the square root function. For a, we'd like to pick something that is easy to compute f of a, well, what's the number close to 59, that is easy to compute the square root of 64 springs to mind. So let's set a equal to
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64. Since we're trying to compute the square root of 59, we want a plus delta x to be 59. In other words, 64 plus delta x is 59. And so delta x should be negative five, it's fine to have a negative number for delta x. Now plugging into our approximation formula, we have f of 59 is approximately equal to f of 64, plus f prime of 64 times negative five. Since f of x is the square root of x, or in other words, x to the one half power, f prime
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of x is going to be one half x to the minus one half power, or one over two times the square root of x. So f prime of 64 is one over two times the square root of 64, which is 1/16. I can rewrite my red equation to say the square root of 59 is approximately the square root of 64 plus 1/16 times negative five, which is eight minus five sixteenths, or 7.6875. using a calculator, I can get a more exact value of the square to 59. my calculator
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says 7.68114575, up to eight decimal places. Let me draw the picture that goes along with this approximation. We have the square root function, and at x equals 64, we're looking at the tangent line. Our delta x here is a negative five, and gets us down to 59. So we're using the value of our tangent line right here to approximate our actual square root function right here. As you can see from the picture, it looks like the tangent line
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value should be slightly bigger than the actual value. And in fact, that's what we get. Then next example is very similar. We call it the linearization of a function is just the equation for its tangent line. Namely, the linearization at a is f of a plus f prime of A times x minus a. And the approximation principle says that f of x, the function is approximately equal to its linearization. Its tangent line, at least when x is near a. This is basically
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the same formula that we use in the last problem, we're just calling our value x this time instead of a plus delta x. Since we're trying to estimate sine of A value, it makes sense to let our function be sine of x. For a, we want to pick a number that's close to 33 degrees, for which it's easy to calculate sine of that number. Well, sine of 30 degrees is easy to calculate. So let's make a equal to 30 degrees, but let's put it in radians and call it pi over six.
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in calculus, we pretty much always want to use radians for sine and cosine especially when taking derivatives. Since the derivative formula, D sine x dx equals cosine of x only works when x is in radians, our x needs to be 33 degrees, since that's the value, we want to estimate the sine of, we need to multiply by pi over 180 degrees to convert it to radians. So that becomes 11 pi over 60 radians. Let's plug in for F and a first to get the linearization.
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And then we'll plug in for x next. So the linearization of our function is going to be sine of pi over six, plus the derivative of sine at pi over six times x minus pi over six. That is l of x is one half, since sine of pi over six is one half, plus cosine of pi over six times x minus pi over six, cosine of pi over six is the square root of three over two. So this is our equation for the tangent line, or the linearization of sine
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of x at pi over six. Now we know that sine of x is approximately equal to as linearisation, as long as x is near pi over six. So in particular, sine of our 33 degrees in radians, which is 11 pi over 60 is approximately equal to one half plus a square root of three over two times 11 pi over 60 minus pi over six. That simplifies to one half plus the square root of three over two times pi over 60. And now I'm going to cheat a little bit and use my calculator to get a decimal value for this
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of about 0.5453. Now if I use my calculator to find sine of 11 pi over 60, directly, remember, that's the same thing as sine of 33 degrees. my calculator tells me it is 0.5446, approximately. So you can see our approximation using the linearization is very close to the calculators, more accurate value. Notice that in this example, the approximate value based on the linearization is slightly higher than the actual value. And you can see why from a graph of sine.
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The tangent line at pi over six lies slightly above the graph of sine x. Therefore, the approximate value based on the linearization will be slightly bigger than the actual value of sine of 33 degrees. In this video, we use several formulas to express one key idea. The main formulas were the approximation principle, the linear approximation, and the linearization. The key idea is that a differentiable function can be approximated near a value x equals
9:28:17
A by the tangent line at x equals a. The differential is a new vocabulary word wrapped around the familiar concept of approximating a function with its tangent line. This figure should look familiar from the previous video on linear approximation is the same picture. Suppose we have a differentiable function, f of x, and we know the value of f at some x value a. That is, we know the value of f evey, but we don't know the value of f at some nearby
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x value a plus delta x. That is we don't know f of a plus delta x. So we draw the tangent line to f of x at x equals a. And we use the tangent line at a plus delta x as an approximation for the function at a plus delta x. Since the tangent line has slope of f prime of a, the rise of a run is f prime of a. So if this run here is delta x, this rise has to be f prime of A times delta x. So the height of the tangent line here at a plus delta x is
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going to be f of a plus f prime of A times delta x. That's just the height here plus the extra height here. And since we're using that height to approximate In our function, we say that f of a plus delta x is approximately equal to f of a plus f prime of A times delta x equivalently. If I subtract F of A from both sides, I get f of a plus delta x minus f of a is approximately equal to f prime of A times delta x. This equation is just the
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approximation principle that we've seen before. And this is a very slight alteration of it. So there's nothing new yet. But now I'm going to wrap some new notation around this familiar concept. The differential dx is another way of writing delta x, you can think of it as a small change in the value of x. The differential df is defined as f prime of x dx, or equivalently f prime of x delta x. Sometimes this is written as d y instead, but d y just means the same
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thing here as df. Sometimes it's handy to specify the differential add a particular value of x, like a value of x equals a, and this is written df equals f prime of a dx, or f prime of a delta x. Notice that the value of a is not apparent when you just write down d f, or d y. Finally, the change in f, which is written delta f, is defined as f of x plus delta x minus f of x for some value of x, for example, f of a plus delta x minus f of a, this can
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also be written as the change in y. Using these new definitions, we can now rewrite our approximation principle to say, delta f is approximately equal to d f, the change in the function is approximately equal to the differential course this could also be written as the change in Y is approximately equal to d y. In the picture, we can now write d x for the run, and D, F, for the rise of the tangent line. Pause the video and take
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a moment to find delta f in this picture, though delta f is f of a plus delta x minus f of a, so that's this height here. I'll write that as delta f, or delta y. So the approximation principle, written in differential notation, is just saying that the rise of the function, delta f is approximated by the rise in the tangent line df. Let's use the differential and an example. For the function f of x equal to x times ln x. Let's first find the differential
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df. We know that df is equal to f prime of x dx. And f prime of x by the product rule is equal to x times the derivative of ln x, which is one over x plus the derivative of x, which is one times ln x. So in other words, one plus ln x. Therefore, df is equal to one plus ln x times dx. When x equals two, and delta x equals negative point three, well delta x is the same thing as dx, we can just plug in those values and get df is one plus
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ln of two times negative 0.3. as a decimal, that's approximately negative 0.5079. Now delta f is defined as f of x plus delta x minus f of x. So for our function, that's x plus delta x times ln of x plus delta x minus x ln x. Plugging in the given values for x and delta x, we get delta f is two minus 0.3 times ln of two minus 0.3 minus two ln two, which according to my calculator is negative zero. Point 4842. And we see that the change
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in the function between two and two minus point three is closely approximated by the change in the tangent line. As expected. The differential is often used to estimate error, as in this example, suppose that the radius of a sphere is measured as eight centimeters with a possible error of point five centimeters. So the sphere that we measure looks something like this, but the actual sphere might be slightly bigger, or slightly smaller, we want
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to use the differential to estimate the resulting error in computing the volume of the sphere. Well, the volume of a sphere is given by four thirds pi r cubed, where r is the radius. If our radius changes by point five centimeters, our volume will change by substantially more. And that change in volume is the error the resulting error in measuring volume. But instead of computing delta V directly, we're asked to approximate it using the differential.
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So we're going to use the fact that delta V is approximately equal to dv, which is easier to compute. By definition, dv is equal to the derivative of our function, I'll just call that v prime as a function of r times Dr. Now, v prime of r is equal to four pi r squared, just by taking the derivative. And here, we're interested in an R value of eight, and a value of Dr. Same as delta r of 0.5 centimeters. So dv is going to be four pi r squared Dr. And when
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I plug in R and D R, I get four pi times eight squared times 0.5, which is 128 pi, or as a decimal 402.1 centimeters. That's our error estimate, which seems quite a bit bigger than our original error of point five in measuring the radius. Now, the relative error of a function is its error over the original value of the function. So in our case, it's the change in volume over the actual volume. Since we're using the differential instead, we'll compute
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the relative error as dv over V. Now, the volume when r is eight centimeters, is four thirds times pi times eight cubed. And dv, we already saw was four pi times eight squared times 0.5. So dv divided by V is given by this ratio, which simplifies to 0.1875. So an 18.75% relative there. To me, the relative error gives a better sense for the error than the absolute error estimate above. This video introduced the idea of the differential, we
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said that we could think of dx as just being another way of writing delta x, but df represents the rise in the tangent line, and is equal to f prime of x times dx. Whereas delta f is the rise in the actual function F. And that's f of x plus delta x minus f of x. On the picture, dx is the run, df is the rise in the tangent line. And delta f is the rise in the actual function. In the language of differentials, we can restate the approximation
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principle to say that the change in f can be approximated by the differential. In the past, we've encountered limits, like the limit as x goes to two of x minus two over x squared minus four. We can't evaluate this limit just by plugging in two for x, because x minus two goes to zero, and x squared minus four goes to zero as x goes to two. This is known as a zero over zero indeterminate form. It's called indeterminate because we can't tell
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what the limit is going to be just by the fact that the numerator goes to zero and the denominator goes to zero. It depends on how fast the numerator and the denominator are going to zero compared to each other. And the final limit of the quotient could be any number at all, or it could be infinity or it could not even exist. In the past, we've been able to evaluate some limits in zero over zero indeterminate form by using algebraic
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tricks to rewrite the quotients. In this video, we'll introduce lopi talls rule, which is a very powerful technique for evaluating limits and indeterminate forms. A limited of the form the limit as x goes to a of f of x over g of x is called a zero over zero indeterminate form, if the limit as x goes to a of f of x is equal to zero, and the limit as x goes to a of g of x is equal to zero. A limit and this form is called an infinity over infinity
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and determinant form. If the limit as x goes to a of f of x is equal to infinity or minus infinity. And the limit as x goes to a of g of x is equal to infinity or minus infinity. We saw an example of a zero over zero indeterminate form in the introductory slide. One example of a infinity over infinity and determinant form is the limit as x goes to infinity of 3x squared minus 2x plus seven divided by negative 2x squared plus 16. Notice
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that as x goes to infinity, the numerator goes to infinity while the denominator goes to negative infinity. In these definitions of indeterminate form, it's possible for a to be negative infinity or infinity, like it is in this example, but it doesn't have to be loopy. talls rule can be applied when f and g are differentiable functions. And the derivative of g is nonzero in some open interval around a except possibly in a under
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these conditions, if the limit as x goes to a of f of x over g of x is zero over zero or infinity over infinity indeterminant form than the limit as x goes to a of f of x over g of x is the same thing as the limit as x goes to a of f prime of x over g prime of x, provided that the second limit exists, or as plus or minus infinity. Let's look at loopy tiles rule in action. In this example, as x goes to infinity, the numerator x goes
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to infinity and the denominator three to the x also goes to infinity. So we have an infinity over infinity indeterminate form. So let's try applying lopi tiles rule, our original limit should equal the limit as x goes to infinity of the derivative of the numerator, which is one divided by the derivative of the denominator, which is ln of three times three to the x, provided that the second limit exists or as infinity or negative infinity.
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In the second limit, the numerators just fixed at one. And the nominator goes to infinity as x goes to infinity. Therefore, the second limit is just zero. And so the original limit evaluates to zero as well. In this example, we have a zero over zero indeterminate form, because as x goes to zero, sine of x and x, both go to zero in the numerator, and sine of x cubed goes to zero in the denominator. So using low Patel's rule, I'll try to evaluate
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instead, the limit I get by taking the derivative of the numerator and the derivative of the denominator, the derivative of sine x minus x is cosine of x minus one, and the derivative of sine x cubed is three times sine x squared times cosine x using the chain rule. Now let me try to evaluate the limit again, as x goes to zero, cosine of x goes to one. So the numerator here goes to zero. As x goes to zero, sine of x goes to zero and cosine of x goes to
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one, so the denominator also goes to zero. So I still have a zero over zero indeterminate form. And I might as well try applying loopy toss rule again. But before I do, I want to point out that cosine of x is going to one. So the cosine of x here really isn't affecting my limit. And in fact, I could rewrite my limit of a product as a product of limits where the second limit is just one and can be ignored from here on out. Now apply lopatok
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rule on this first limit, which is a little bit easier to take the derivatives and so the derivative of the top is minus sine x. And the derivative of the bottom is six times sine x times cosine x. Now let's try to evaluate again, as x goes to zero, our numerator is going to zero, and our denominator is also going to zero. But hang on, we don't have to apply lobby towels rule again, because we can actually just simplify our expression,
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the sine x on the top cancels with the sine x on the bottom. And we can just rewrite our limit as the limit of negative one over six cosine of x, which evaluates easily to negative one six. In this example, I want to emphasize that it's a good idea to simplify after each application of lopi talls rule. If you don't simplify, like we did here, then you might be tempted to apply loopy towels rule and additional time when you don't need to, which might make
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the problem more complicated. Instead of simpler to solve this video, we were able to evaluate zero over zero and infinity over infinity indeterminate forms by replacing the limit of f of x over g of x with the limit of f prime of x over g prime of x, provided that second limit exists. This trick is known as lopi tels rule. We've seen that lopatok rule can be used to evaluate limits of the form zero over zero, or infinity over infinity.
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In this video, we'll continue to use lopi towels rule to evaluate additional indeterminate forms, like zero times infinity, infinity to the 00 to the zero, and one to the infinity. In this example, we want to evaluate the limit of a product. Notice that as x goes to zero from the positive side, sine x goes to zero, and ln x goes to negative infinity. Remember the graph of y equals ln x. So this is actually a zero times infinity indeterminate form.
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Even though the second factor is going to negative infinity, we still call it a zero times infinity and indeterminate form, you can think of the Infinity here as standing for either positive or negative infinity. It it's indeterminant. Because as x goes to zero, the sine x factor is pulling the product towards zero, while the ln x factor is pulling the product towards large negative numbers. And it's hard to predict what the limit of
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the product will actually be. But the great thing is, I can actually rewrite this product to look like an infinity over infinity and determinant form, or a zero over zero and determinant form. Instead of sine x times ln x, I can rewrite the limit as ln x divided by one over sine x. Now as x goes to zero, my numerator is going to negative infinity. And since sine x is going to zero through positive numbers, my denominator one over
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sine x is going to positive infinity. So I have an infinity over infinity indeterminate form. Now, I could instead choose to leave the sine x in the numerator, and instead, put a one over ln x in the denominator. If I do this, then as x goes to zero through positive numbers, sine x goes to zero. And since ln x goes to negative infinity, one over ln x goes to zero. And so I have a zero over zero indeterminate form. Sometimes it
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can be difficult to decide which of these two ways to rewrite a product as a quotient. One rule of thumb is to take the version that makes it easier to take the derivative of the numerator and denominator. Another trick is just to try one of the ways and if you get stuck, go back and try the other. I'm going to use the first method of rewriting it because I recognize that one over sine x can be written as cosecant of x. And I know
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how to take the derivative of cosecant x. Using low Beatles rule on this infinity over infinity and determinant form, I can rewrite my limit as the limit of what I get when I take the derivative of the numerator, that's whenever x divided by the derivative of the denominator, that's negative cosecant x cotangent x. As always, I want to simplify my expression before going any further. I can rewrite my trig functions in the denominator in terms
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of sine and cosine. cosecant x is one over sine x cotangent x is cosine of x over sine of x. Now flipping and multiplying, I get the limit as x goes to zero plus of one over x times sine squared of x over negative cosine of x. In other words, the limit of negative sine squared x over x cosine x, we know that cosine of x goes to one as x goes to zero. So I can rewrite this as the limit of negative sine squared x over x times the limit of something that goes to one. So
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I once again have a zero over zero indeterminate form. And I can apply lopatok rule yet again, taking the derivative of the top, I get negative two, sine x, cosine of x. And the job of the bottom is just one. Now I'm in a good position just to evaluate the limit by plugging in zero for x in the numerator, I get negative two times zero times one, the denominator is just one, so my final limit is zero. In this limit, we have a battle of forces. As
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x is going to infinity, one over x is going to zero. So one plus one over x is going to one, but the exponent x is going to infinity, it's hard to tell what's going to happen here. If we had one, to any finite number, that would be one. But anything slightly bigger than one, as we raise it to a bigger and bigger powers, we would expect to get infinity. So our limit has an independent permanent form, it's hard to tell whether the answer is going
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to be one infinity, or maybe something in between. Whenever I see an expression with a variable in the base, and a variable in the exponent, I'm tempted to use logarithms. If we set y equal to one plus one over x to the x, then if I take the natural log of both sides, I can use my log roles to rewrite that by multiplying by x in the front. Now, if I wanted to take the limit as x goes to infinity of ln y, that would be the limit of this product,
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x times ln one plus one over x. As x goes to infinity, the first factor x goes to infinity. One plus one over x goes to just one and ln one is going to zero. So we have a infinity times zero indeterminate form, which we can try to rewrite as an infinity over infinity, or a zero over zero indeterminate form. Let's rewrite this as the limit of ln one plus one over x divided by one over x. This is indeed a zero over zero in determinant form. So we
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can use lobi tiles rule and take the derivative of the top and the bottom, the derivative of the top is one over one plus one over x times the derivative of the inside, which would be negative one over x squared. And the derivative on the bottom, the derivative of one over x is negative one over x squared, we can actually cancel out these two factors, and rewrite our limit as the limit as x goes to infinity of one over one plus one over
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x, which is just equal to one, since one over x is going to zero. So we found that the limit of ln y is equal to one, but we're really interested in finding the limit of y, which we can think of as e to the ln y. Since ln y is going to one, e to the ln y must be going to e to the one. In other words, E. So we found that our original limit is equal to E. And in fact, you may recognize that this limit is one of the ways of defining IE.
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In the previous example, we had a one to the internet Today in determinant form, and we took logs and use log roles to write that as an infinity times zero and determinant form. Well, the same thing can be done if we have an infinity to the zero indeterminate form, or a zero to the zero indeterminate form. So one to the infinity, infinity to the zero, and zero to the zero, are all indeterminate forms that can be handled using lobi toss
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rule. In this video, we saw that a zero times infinity indeterminate form could be converted to a zero over zero, or infinity over infinity indeterminate form by rewriting f of x times g of x as f of x divided by one over g of x, or as g of x divided by one over f of x. We also saw how to use lopi talls rule on these three sorts of indeterminate forms by taking the ln of y, where y is our f of x to the g of x that we want to take the limit
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of. This video is about Newton's method for finding the zeros of a function, f of x. In other words, the values of x that make f of x equal to zero. The zeros of a function can also be thought of as the x intercepts of its graph. Suppose we want to find a solution to the equation either the x equals 4x. This equation cannot be solved using standard algebraic methods. For example, taking the ln of both sides doesn't really help because we still
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get x equals ln of 4x, which is just as hard to solve. Instead, we can look for approximate solutions. Looking at the graph of y equals e to the x and y equals 4x, we see there should be two solutions, one at approximately x equals little more than two and the other around x equals maybe point three or point four. Newton's method will allow us to make much more accurate approximations to the solution of this equation, then we can do by just glancing
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at the graph. To use Newton's method, instead of looking at the equation, e to the x equals 4x. We'll look at the equation E to the X minus 4x equals zero. And in fact, we'll define the function f of x to be e to the x minus 4x. And look for zeros of that function. After all, finding a zero of this function is the same as finding a solution to our original equation. So now we're trying to solve the equivalent problem of finding the zero of
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the function f of x equals e to the x minus 4x. That's the function that's drawn below. I'm going to focus on this zero, the one near to, and I'm just going to make an initial guess anything reasonably close to the actual zero should do. So I'll just put an initial guess right here, and I'll call it x one. Now x one is not actually zero of my function, and I'll write the point on the graph above it as x one, f of x one. To get a better estimate
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for the zero of my function, I'm going to make use of the tangent line to my function that goes through this point. So the second step will be to find this tangent line. Since the tangent line is a reasonably good approximation to the function, the point where the tangent line crosses the x axis should be closer to the point where the function itself crosses the x axis, which is the point I'm looking for. So the third step will be to find the
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x intercept for the tangent line. I'll call this x intercept, x two. Now I'm just going to repeat this process. I'll use x two as my next guess. I'll follow it up to the function where I have the point x to f of x two, and then I'll draw a new tangent line and get a new intercept. I can repeat this process as often as I need to, to get a sufficiently accurate approximation to my actual zero of my function. Now that I've described the process
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graphically, let's find some equations that go along with this picture. If I start with the initial guess, of x one, then the tangent line through x one, f of x one is given by the x Bayesian y equals f of x one plus f prime of x one times x minus x one. You might remember this equation from a section on linearization. And it's really just comes from the formula y minus y one equals m x minus x one that holds for any line, where
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m here is the derivative at x one, and y one is f of x one. plugging into that equation, we have y minus f of x one equals f prime of x one times x minus x one, which simplifies to y equals f of x one plus f prime of x one times x minus x one. So that's where this linearization equation comes from. It's just the equation of the tangent line. Now, if we want to find the x intercept of the tangent line, we just said
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the tangent line equation equal to zeros, we have zero equals f of x one plus f prime of x one times x minus x one, and we solve for x. so this can subtract f of x one from each side, divided by f prime of x one and solve for x. We're calling this new x intercept x two. So x two is x one minus f of x one over f prime of x one. Now we have our second guess, x two, and we can again find the tangent line through x two, f of x two, that tangent
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line will be given by the same sort of equation. And if we then find the x intercept, the same algebraic steps, get us to the analogous equation x three equals x two minus f of x two over f prime of x two. And more generally, as we repeat this process over and over again, our n plus one guess is going to be given by x n plus one equals x n minus f of x n over f prime of x n. That's the J equation at the core of Newton's method. Now that we've got
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the theory down, let's grind through the problem at hand with some numbers. Our function has the equation f of x equals e to the x minus 4x. So f prime of x is e to the x minus four. So from Newton's methods equation, we have in general, X sub n plus one is X sub n minus e to the x sub n minus four times X sub n over e to the x sub n minus four. Let's start with for example, x sub one equals three, then x sub two is going to be three minus
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e cubed minus four times three over e cubed minus four. plugging this into a calculator, I get x sub two equals 2.4973, and so on. Now to cube compute x sub three, I have to take this whole number and plug that in to my formula. I've written it out as just 2.49. But for accuracy, when I actually computed my calculator, I'll use the entire number. my calculator gives me this answer for x of three and continue in this process, I can
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get x of 4x 5x sub six. If I compute one more, x of seven, I noticed that I have no change to my value in the number of digits that the calculator spits out. So at this point, my Newton's methods iterations have converged. And I have an answer that's accurate to about eight decimal places. I found one zero for my function. And if I wanted to find the second zero, the one over here, I would just need to start with an initial value that's close
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to this x coordinate, perhaps an initial value of zero might be good. In this video, we developed an algorithm for getting increasingly accurate approximations to the zero of a function. The central equation that we used was this one which tells us how to get from one approximation, X sub n to the next 1x sub n plus one. When we go from a function, say 3x plus sine x to its derivative, in this case, three plus cosine x, that's called differentiating, or
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finding a derivative. Anti differentiating, or finding an antiderivative, takes us the other direction, from a derivative to a function that has that as its derivative. For example, if g prime of x is 3x squared, that's the derivative. What could g of x the original function be? Well, g of x could be x cubed, since the derivative of x cubed is 3x squared. Or it could also be g of x equals x cubed plus seven, for example, or g of x
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equals x cubed plus any constant, where I write a general constant with a capital C. That's because the derivative of a constant zero, so the derivative of x cubed plus a constant is just going to be 3x squared, no matter what the constant is. A function capital F of X is called an antiderivative of lowercase F of X on an interval a, b, if the derivative, capital F prime of X is equal to lowercase F of X on that interval a b. In other words,
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we can think of little f as being the derivative of the function capital F. In the above example, x cubed is an antiderivative of 3x squared. And in fact, x cubed plus C for any constant C is also an antiderivative of 3x squared. When we add on a general constancy, that's sometimes referred to as a general antiderivative, we found a general family of anti derivatives for the function 3x squared. But could there be other anti derivatives, other functions
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whose derivative is 3x squared. In fact, there are no others. And one way to think about this intuitively, is if you have two functions with the same derivative, it's like having two runners in a race that always speed up and slow down at exactly the same times. If one of those runners starts ahead of the other, then the distance between them will always stay exactly the same. That's the vertical distance drawn here on the graph. And that's
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the constant C, that separates one antiderivative y equals x cubed from another y equals x cubed plus C. And in general, if capital F of X is an antiderivative for a little f of x, then all other anti derivatives can be written in the form capital F of x plus C for some constancy. A more rigorous justification of this fact, can be proved using the mean value theorem, as I'll do in a separate video. If you know the derivatives for some standard
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functions, then it's pretty easy to get some anti derivatives. For example, the antiderivative of one is x. Since the derivative of x is one, if we want to make that a general antiderivative, we can add a constant C, the antiderivative of x is x squared over two because when I take the derivative of x squared over two, the two that I pulled down and multiply cancels with the two in the denominator, leaving me x. Again, I can make this a more general antiderivative
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by adding a constant C. More generally, the antiderivative of x to the n for any and that's not equal to negative one is given by x to the n plus one divided by n plus one plus a constant C. I can check this by taking the derivative of x to the n plus one over n plus one. The n here is just a constant. So using the power rule, I get n plus one times x to the n divided by n plus one that yields x to the n, which is what I want it. We can
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think of this rule as the power rule for anti differentiating since it's closely related to the power rule for differentiating. Now, this rule doesn't apply when n equals negative one. Notice that we'd be dividing by zero if n were negative one but we can handle the case when n equals negative one separately, since x to the negative one is one of our x, we recognize that the antiderivative of one of our x is just ln of the absolute value
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of x plus C. Since the derivative of ln of the absolute value of x is one over x, please pause the video and see how many more anti derivatives you can fill in in this table. You should get all of these formulas based on the analogous formulas for differentiating, notice that the antiderivative of sine x is negative cosine x not cosine x, because the derivative of cosine x is negative sine x. If I have a constant times x to the n, I'm
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going to call the constant a instead of C, since I've already got some C's floating around. If I want the antiderivative of A times x to the n, that's just going to be a times the antiderivative of x to the n, which is x to the n plus one over n plus one plus a constant say that's because when I take the derivative of a constant times a function, I can just pull the constant out. More generally, the antiderivative of a constant A times any
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function, little f of x is just going to equal A times the antiderivative of little f of x, which I'll denote with capital F of X, plus a constant C. The antiderivative of f of x plus g of x is capital F of X, plus capital G of x plus c, where capital F and capital G are the antiderivative of lowercase F and lowercase J. This is because the derivative of a sum is equal to the sum of the derivatives. Let's use this information to compute the
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antiderivative for f of x equals five over one plus x squared minus one over two times the square root of x. First, I'm going to rewrite f of x as five times one over one plus x squared minus one half times x to the minus one half. I know that the antiderivative of one over one plus x squared is arctangent of x. And by the power rule for anti differentiating the antiderivative of x to the minus one half, I get by raising the exponent by one, negative
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one half plus one is positive one half, and then dividing by the new exponent by my constant multiplication rules, I can just multiply by my constants. And that's my antiderivative capital F of X, I've have to remember the plus C for the general antiderivative, I can simplify a little bit by canceling these one halves. And I get a final answer of five times arc tan of x minus a squared of x plus C. In this video, we introduced anti derivatives
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and build a table of anti derivatives based on our knowledge of derivatives. In this video, we'll solve problems where we're given an equation for the derivative of the function. And we're given an initial condition, something like f of one equals seven. And we have to find the function f of x. In this first example, suppose g prime of x is e to the x minus three times sine x. and g of two pi is five, we need to find g of x. Well, g of x is an antiderivative
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of e to the x minus three sine x. So g of x is of the form e to the x plus three cosine of x plus a constant C. That's because the derivative of e to the x is e to the x, the derivative of cosine x is minus sine x, and the derivative of a constant is just zero. Now I need to find the value of the constant C that makes this initial condition come true. If I plug in two pi for x, I get e to the two pi plus three times cosine of two pi plus
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C, and that needs to equal five. Since cosine of two pi is one, I have that e to the two pi plus three plus c equals five. And so C is equal to two minus e to the two pi. So my function g of x is equal to either the x plus three cosine x Plus two minus e to the two pi. In this example, we're given the second derivative of f. And we're given two initial conditions, f of one is zero, and f of zero is two.
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To start, I'm going to rewrite f double prime of x in a more manageable form. by distributing, I get the square root of x times x minus the square root of x over x. and rewriting with exponents, we get x to the three halves minus x to the minus one half. Next, I'm going to find f prime of x, which is the antiderivative of f double prime of x. So f prime of x, using the power rule for anti derivatives, I raised the exponent of three halves by one to get
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five halves, and then divide by the new exponent five halves. Similarly, I raised negative one half by one to get one half and divide by the new exponent one half. And I'll add on a constant see, let me simplify a little here, instead of dividing by five halves, multiply by two fifths, and instead of dividing by one half, multiply by two. Now I've got an expression for f prime of x, but I need an expression for f of x, which is the antiderivative
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of f prime, and so all anti differentiate again. So now I have two fifths times x to the seven halves over seven halves minus two times x to the three halves over three halves, the antiderivative of a constant C is C times x, and then I'll add on a new constant D. After simplifying a bit, I'm ready to use my initial conditions, in order to solve for my constant C and D. When I plug in zero for x, all my x terms vanish, I'm just left with
9:34:44
D, so d has to equal two. So I can rewrite my function, setting D equal to two. And now my second condition says that f of one equals zero. So plugging in one for x, I get 430 fifths minus four thirds plus c plus two, and that has to equal zero, which means that c is negative two, minus 430 fifths, plus four thirds, which simplifies to minus 80 to 100, and fifths. If we plug that in for C, we get a final answer for f of x. And that
9:34:49
finishes the problem. In this final example, we're not given any equations, so we have to make them up ourself, we're told that we're standing at the edge of a cliff, at height 30 meters, with throw a tomato up in the air at an initial velocity of 20 meters per second. The tomato then falls down to the ground due to gravity. And we want to find how long that takes and what its velocity is at impact. We know that the acceleration due to gravity
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is negative 9.8 meters per second squared. If we're working in metric units. The similar figure if we're working in units of feet, is negative 32 feet per second squared. The negative sign is because gravity is pulling objects down towards the ground in the negative direction. We're also given the initial condition, that velocity at time zero is 20 meters per second. That's a positive velocity because we're throwing the tomato up. And we're told
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that the initial position s of zero is 30 meters. So let's start with the equation that acceleration is negative 9.8. In other words, S double prime of t is negative 9.8. Therefore, S prime of t is negative 9.8 t plus a constant c one. And from my initial condition about velocity, I know that s prime of zero is 20. So in other words, S prime of zero is negative 9.8 times zero plus c one that has to equal 20. Which means that c one is equal to 20.
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Substituting in 20 for C one, I can rewrite s prime of t Now I can find f of t, the antiderivative of S prime. And that's going to be negative 9.8 times t squared over two plus 20 t plus a second constant C two. Using my second initial condition, f of zero equals 30. I can plug in zero for t, and get an expression that equals 30. Since all the terms drop out, besides the C two, that tells me that C two is 30. And so I can find my equation position by substituting 34 c two.
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Now I want to find out how long it takes the tomato to hit the ground. So that's going to be the time when s of t equals zero. Setting zero equal to my expression for S of t, I can use the quadratic formula to solve for t, and I get t equals approximately negative 1.17, or 5.25. The negative time doesn't make sense for the problem. So I'm left with a time of impact of 5.25 seconds. Now to find the velocity of the impact, I need to plug
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that time into my velocity equation. In other words, my equation for S prime. So s prime of 5.25, is 9.8 times 5.25, plus 20, which simplifies to negative 31.45 meters per second, probably enough to squash the tomato. And that's all for this video on finding anti derivatives using initial conditions. In this video, I'll use the mean value theorem to show that the antiderivative zero has to be a constant. And then any two anti derivatives
9:35:20
of the same function have to differ by a constant. In a previous video, I stated the fact that if f of x is one antiderivative of a function, little f of x, than any other antiderivative of that same function can be written in the form capital F of x plus C for some constant C. In other words, any two antiderivative of the same function have to differ by a constant. To prove this fact, let's first note that if the derivative of a function, g prime of
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x is equal to zero on an interval than the original function, g of x must equal C for some constant C. This statement follows from the mean value theorem, because the mean value theorem tells us that for any x one and x two in our interval, the average rate of change between x one and x two is equal to the derivative at some number x three in between x one and x two. But by assumption, g prime is zero everywhere on the interval, so g prime of
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x three must be equal to zero. This means that our numerator, g of x want to minus g of x one has to equal zero. In other words, g of x two is equal to g of x one, but that's true for any x one and x two. So all the values of G are equal, and G must be a constant. The second observation that I want to make is that if G one and G two are two functions, which have the same derivative, then g one of x must equal g two of x plus C for some
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constant C. This statement follows from the previous statement, because if G one prime of x is equal to g two prime of x, then g one prime of x minus g two prime of x must equal zero. But that means if I look at the function g one of x minus g two of x, and take its derivative, that has to equal zero, since the derivative The difference is the difference of the derivatives. Now our previous statement tells us that if the derivative
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of a function is zero, the function must be a constant, and therefore, g one of x minus g two of x equals C for some constant C, which means that g one of x is equal to g two of x plus C, which is what we wanted to prove. So we've proved that any two functions with the same derivative have to differ by a constant or in other words, If capital F of X is one antiderivative of a function than any other antiderivative must be of the form capital
9:35:45
F of x plus C. This concludes the proof that any two anti derivatives of a particular function must differ by a constant. This video will review summation notation, that is, the sigma notation used to write a song. In this expression, using the greek capital letter sigma, the letter I is called the index. The number one is called the lower limit of summation, or the starting index. And the number five is called the upper limit of summation, or the ending index, we evaluate this expression
9:35:51
by summing up to the I, for all values of i starting from one and ending at five and stepping through the integers. In other words, we start with i equals one and take two to the one. And then we add to it two to the two, two to the Three, two to the four, and two to the five. If we do the arithmetic, this comes out to 62. In the second expression, our index is J, and we start with J equals three and go up to J equals seven, once again,
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stepping through integer values. So we have to add up 1/3 plus 1/4, plus 1/5, plus one, six, and then our last term is one seven. This sum is equal to 153, over 140. When we're given a psalm like this one, it can be handy to write it in sigma notation, because it's more compact. But to do so, we have to look for the pattern between the terms. In this case, the terms all differ by three. So I can think of nine as being six plus three,
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and 12, as being six plus twice, three, and 15 as being six plus three times three, and so on. In fact, we can even think of six itself as being six plus zero times three to fit this pattern. Now we can write the sum as sigma of six plus i times three, where I ranges from zero to four. Here, we're thinking of six plus three as six plus one times three. Now, there are other ways to write this sum and sigma notation. For example, we could
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notice that each of the terms is a multiple of three. And in fact, six is three times two, nine is three times three, and so on. And so we could write our sum as sigma of three times n, where n ranges from two to six. The choice of the letter we use for the index doesn't matter at all. For example, we could also write this as sigma from K equals two to six of three times k. Here, k and n play the same role. Please pause the video
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for a moment and try to write this next example in sigma notation. Since the denominators are powers of two, we could write the denominators as two to the i, where i ranges from two to five. The numerators are one less than the denominators, so we can write the numerators as two to the i minus one was we're adding these terms up, we write sigma, and we I go from two to five. In this video, we reviewed summation notation, or sigma notation for
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writing sums. In this video, we'll approximate the area under a curve using tall skinny rectangles, this will introduce the idea of an integral. Let's start by approximating the area under this curve y equals x squared in between x equals zero, and x equals three by approximating it with six rectangles. There's several ways to do this. For example, I could draw rectangles for the right side of each rectangle is as tall as the curve. We'll call this using write endpoints. Alternatively,
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we could line up our rectangles, so the left side of each rectangle is as high as the curve. We'll call this use Left endpoints. Notice that the leftmost rectangle in this picture is degenerate and has height zero. If we're using the picture with the right endpoints, then the base of each rectangle has size one half. And the height of each rectangle is given by the value of our function, y equals x squared on the right side of the rectangle.
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So for example, the area of the first rectangle is its base times its height, which is point five times 0.5 squared. The 0.5 squared comes from me evaluating the function at this point 0.5 to get the height. Similarly, the area of the second rectangle is going to be also base times height basis, still point five, and now the height is going to be one squared, or one. If we continue like this, and add up all our areas, we get the area of all six
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rectangles is given by this expression. Notice that there are six terms here, one for each rectangle, each rectangle has base of point five, and has height given by the right endpoint, we can write this in sigma notation as the sum of 0.5 times 0.5 times i squared, where I ranges from one to six. This works because the numbers in parentheses here are all multiples of point five. The first one is point five times one, and the last one, three is point
9:36:37
five times six. Now if we compute the sum, we get 91 eighths, which is 11.375. Notice from the picture, that the sum of areas of rectangles is an over estimate for the area under the curve, we can do the same sort of computation for this green picture using left endpoints, and we'll get an under estimate for the area under the curve, I invite you to try it for yourself before going on with the video. For the green rectangles, the first
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rectangle has area zero, the second rectangle has area given by its base of 0.5 times its height of 0.5 squared. And if we compute all six areas and add them up, we get a similar expression to the previous one, only this time, we end with a 2.5 squared, which is the height of our last rectangle. One way to write this in sigma notation, is still starting with i equals one for the first rectangle to six for the last one, we use the base. And then for the height, we use 0.5 times I minus one,
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squared. This works because when I is one, i minus one is zero, so we start with a height of zero like we should. And when I is six, we get point five times six minus one, which is point five times five, or 2.5, just like we want it to be, if we add up the sum, we get an answer of 55 eighths, which is equal to 6.875, which is an underestimate for the true area under the curve. Now there's a big gap between 6.875 and 11.375. So it'd be nice
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to get tighter bounds on the area. One way to do this is by using more rectangles, for example, 12 rectangles instead of six. Again, we could choose to use right endpoints, which gives us an over estimate of area in this case, or we could use left endpoints, which gives us an under estimate. The area of the eyes rectangle is given by the base times the height, and the base is going to be in this case 0.25. While the height is given
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by the functions value on the right endpoint, the right endpoint of the eyes rectangle is going to be 0.25 times I and the function is the squaring function. So that height will be given by point two, five i squared. The area of all the rectangles can then be given by the sum from i equals one to 12 for the 12 rectangles of 0.25 times 0.25 i squared. If we work out that sum, it comes to 10.156. Again, we can do the same thing with left
9:37:05
endpoints. Now the area of the eye, the rectangle, is given by base times height, which is point two, five. And now the height is given by the value of the function on the left endpoint. So that's going to be 0.25 times I minus one. And we need to square that, since our function y equals x squared is giving the height of my rectangles. So the area of all the rectangles together is going to be the sum from the first rectangle to the 12th rectangle of 0.25 times 0.25, times I minus one squared.
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That works out to 7.906. So we're honing in on the actual area under the curve. Now it's somewhere between about eight and about 10, we can keep getting better and better estimates of area by using more and more rectangles. For example, if we want to use 100 rectangles, then our area of all rectangles using right endpoints is going to be given by the sum from i equals one to 100 of the basis times the heights. Now the base of each rectangle
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will be 100 of the length here from zero to three. In other words, the base will be three over 100. The eyes right endpoint, which I'll call X sub i, is going to be just three over 100 times I, since you get to the right eye threat endpoint by taking I copies of a rectangle of width three one hundredths. Therefore, the eyes height is going to be given by this is right endpoint squared, or three one hundredths times i squared. So we can write our sum of
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areas as sigma from i equals one to 100 of three one hundredths times three one hundredths times i squared. The formula using left endpoints is similar. The if left endpoint is going to be three one hundredths, times I minus one, since if we're using left endpoints, we only have to travel through i minus one rectangles to get to the left endpoint of the rectangle. So our area for the left endpoints becomes the sum from i equals one to 100 of
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three over 100 times three over 100 times I minus one squared. These two psalms work out to be 9.1435 and 8.8654. At this point, you might be willing to wager a guess that our exact area under the curve is probably going to be nine. But to determine the exact area for sure, let's do this process of dividing into rectangles one more time. And this time, we'll just use an rectangles where n is some big number. Since we're dividing an interval
9:37:32
of length three into n little pieces, the width of each sub interval, in other words, the base of each rectangle, we'll be given by three over n. I'll call this delta x as a little tiny bit of x. Now the right endpoint, x by is given by three over n times i. Since we have to travel through I rectangles, each of width three over n on order to get to that right endpoint. So our height, H sub i is given by the functions value on that right
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endpoint. We can work out similar expressions for the picture using left endpoints here. Our estimate of area using right endpoints is then the sum from i equals one to n of three over n times three I over n squared. And our estimate using left endpoints is a sum from i equals one to n of three over n times three, i minus one over n squared, the more rectangles we use, in other words, the bigger the value of n, the closer our estimated
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area will be to our exact area under the curve. And therefore, the exact area is given by the limit the limit as n goes to infinity of this song, which is known as a Riemann sum, there are really two possible limits, we could use right endpoints or left endpoints. But as the picture suggests, these two limits should turn out to be the same thing. In fact, there are other options between sides using right endpoints and left endpoints, we could,
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for example, use the midpoints of our intervals to compute our areas of rectangles. And that limit should also end up as the same thing. So we have an expression for the area under the curve y equals x squared between the values of x equals one and x equals three. And that's given by the limit of this psalm called a Riemann sum. I'll stick with the right endpoint version for now, to compute the exact area, we have to evaluate this limit, which is tricky. I'm going to start by rewriting.
9:37:55
Since three and n don't involve the index I, I can pull them outside of this summation sign. I'll clean this up a little bit. Now will you need to use the fact that the sum of the first n squares of the integers is equal to n times n plus one times two n plus one over six, we can check that formula for a few values of n. For example, if n equals two, we're summing out one squared plus two squared, which is five. And we're plugging in two times
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two plus one times four plus one over six from the formula, which also equals five. If we use this formula, in our limit calculation, we get this expression, which simplifies to nine halves, by dividing 27 by six, we can cancel a copy of n and get n plus one times two n plus one over n squared. I'm going to pull out the nine halves. And now I observe I have the limit of a rational expression, where the highest power term on the numerator
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is going to be two n squared, the highest power and there's nominators just the n squared, so that's going to be a limit of two, multiply that by by nine halves, and I get a limit of nine, just like I expected from the previous work. So that was a big production. But we did successfully find the area under the curve, and it was nine. In this video, we approximated the area under a curve by taking the limit as the number of rectangles goes to infinity
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of the area of the rectangles, which is given by a psalm called a Riemann sum of the basis times the heights of the rectangles. The basis are often written as delta x, and the heights are given by the function value on the left endpoint, or the right endpoint, or some other point in the interval. For our purposes, f was always x squared. But this sort of expression, called a Riemann sum, can be used more generally, to evaluate the area under any continuous
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function. In previous sections, we thought of the definite integral as representing area, and we've computed it as a number. In this section, we'll think of the integral itself as a function of the bounds of integration. And we'll describe the first part of the fundamental theorem relating the derivative and the integral. Suppose f of x has the graph shown here, and let g of x be the integral from one to x of f of t dt. I'm using t as my variable inside
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my integrand Here, just to distinguish it from the variable x that I'm using in my bounds of integration, this expression just means the net area between one and some value x on the x axis. I'll call geovax, the accumulated area function, because as x increases, g of x, measures how much net area has accumulated. Let's calculate and plot some values of g of x. g of one is the integral from one to one of f of t, dt, that's just zero. Since
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the bounds of integration here the same, g of two is the integral from one to two of f of t, dt. That's the net area from one to two, which is to square units. g of three is the integral from one to three. Now, we've added on an additional two units here, and an additional one unit up here from this triangle, for a total of five, g of four is g of three with some additional area tacked on the additional area measures three units. So g of four is eight. Please pause the video and fill in
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the next few values of J. When we go from g of four to g of five, we add on an extra unit of area. So g of five is nine. As we go from g of five to g of six, we start accumulating negative area, because f is now below the x axis. So here I've accumulated one unit of negative area, which means that g of six is one less than g of five. In other words, gf six is eight, g of seven is five. Since we accumulate three more units of negative area, to find g of zero, the integral from
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one to zero of f of t dt, I'm going to rewrite this integral as negative the integral from zero to one of f of t dt. Since there are two units of area between zero and one, g of zero is negative to apply all these values of G on these coordinate axes, and connect the dots to get an idea of what g of x looks like. Now let's think about the derivative g prime of x. We know that g prime of x is positive, where g of x is increasing, but
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g of x is increasing, wherever we're adding on positive area, that is when f of x is positive. So we have that g prime of x is positive, where f is x is positive. Also, g prime of x is negative, where g of x is decreasing. That happens when we're adding on negative area because f of x is negative. So we can see that g prime of x is negative, where f of x is negative. Also, g prime is zero at this local maximum, where f is zero. At that instant, we're not adding on any positive or negative area.
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If we look a little closer, we can see the rate at which g of x is increasing depends on the height of f of x. When f of x is tall, or high, we're adding on area very quickly. While when f of x is low or small, we're adding on area more slowly. So the rate of change of G. In other words, g prime of x is behaving very much like the function f of x itself. And in fact, it turns out that g prime of x is equal to f of x. This is the first part of the fundamental theorem of calculus.
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The Fundamental Theorem of Calculus Part One says that of f of x is a continuous function on the closed interval from a to b, then for any x in this interval, the function g of x, the integral from a to x of f of t dt is continuous on the interval a b and differentiable on the inside of this interval, and for Furthermore, g prime of x is equal to f of x, as we saw in the previous example. The proof of this fact relies on a limit definition of derivative,
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and can be found in a later video. For now, let's do some examples based on this fact. First, let's find the derivative with respect to x of the integral from five to x of the square root of t squared plus three dt. The Fundamental Theorem of Calculus tells us that this expression here thought of as a function of x is differentiable, and its derivative is just the integrand function evaluated on x. This is great, we don't have to do any
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work here at all. To evaluate the derivative, we just plug in x, where we see the T here, the derivative and the second expression is also the square root of x squared plus three, it might seem odd that these two expressions have the same derivative. But remember, in both cases, we're taking the derivative of the accumulated area function. And the rate at which area accumulates doesn't depend on the lower bound of the integral. That is,
9:39:13
it doesn't depend on where we start counting, it just depends on the height of the function at x. For this third example, remember that the integral from x to four is the same thing as the negative of the integral from four to x. So we get the negative of the derivative from four to x. and applying the fundamental theorem of calculus, this lesson is negative the square root of x squared plus three, it makes sense that we should get a negative
9:39:19
answer for this example. When we're integrating from X to four, then as x increases, our area actually decreases. So our accumulated area function should have a negative derivative. This last example is more complicated, because instead of just having x as our upper bound, we have a function of x sine of x, we can think of sine of x as being the inside function, and the accumulated area function as being an outside function and apply the chain rule.
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In general, the chain rule says that we have the derivative with respect to x of a function of u of x, then that's the same thing as the derivative with respect to u of that function at U times the derivative of the inside function u of x with respect to x, applying the chain rule to our accumulated area function, where you have x is sine x, we have that the derivative respect to x of the integral from four to sine x of the integral of t squared plus three
9:39:29
DT can be written as the derivative by two you, I've accumulated area function from four to you have the integrand times the derivative with respect to x of our u of x, which is just sine x, we can apply the fundamental theorem of calculus to calculate the first derivative. By just plugging in you for T, we get you squared plus three. And then the derivative of sine x, of course, is just cosine of x. Since we want our final answer to be
9:39:34
entirely in terms of x, we're going to rewrite this as the square root of sine x squared plus three times cosine x, or just the square root of sine squared x plus three times cosine of x. We could have gotten this answer more quickly by just plugging in this entire expression sine of x in where we saw the T here in the integrand, and then multiplying the answer by the derivative of sine x due to the chain rule.
9:39:39
This video introduced the fundamental theorem of calculus part one that says that the derivative of the integral of a function is just the original function in some sense, taking the derivative and does the process of taking the integral. derivatives and integrals are closely related. inverse operations. This video introduces the second part of the fundamental theorem of calculus. Another way of relating derivatives and integrals. Part Two of the
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fundamental theorem of calculus says that if f is a continuous function on the closed interval a b, then the integral from a to b of f of x dx is equal to capital F of b minus capital F of A, where capital F is any antiderivative for lowercase F. That is, capital F is a function whose derivative is lowercase F. The proof of part two of the fundamental theorem of calculus follows directly from part one. And I'll give that proof in another
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video. But here, I just want to make a few comments about what this theorem means. If we think of f of x as the derivative of capital F of x, then this is saying that the integral of the derivative is equal to the original function evaluated on the endpoints. I also want to comment on the phrasing any antiderivative. Suppose capital G of X is a different antiderivative for lowercase F. We know that any two anti derivatives differ by a constant. So we know
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that g of x has to equal capital F of X plus some constant. So if we take g of b minus g of a, that's going to be the same thing as f of b plus c minus f of a plus C. And since this constant sees subtract out to cancel here, this is just f of b minus f of a. So this difference is the same value, no matter which antiderivative of lowercase f we use. And that's why we can say that capital F can be any antiderivative. Part Two of the fundamental
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theorem of calculus is super useful, because it allows us to compute integrals simply by finding anti derivatives and evaluating them. Finding anti derivatives tends to be really easy. Computing integrals, using the Riemann sum definition is really hard. And so because of the fundamental theorem of calculus, we don't have to go through all those lengthy and tedious computations, we've involving limits of areas of rectangles, all we have
9:40:04
to do to evaluate an integral is find an antiderivative, and evaluate it. Let's see how this works in some examples. In this first example, the antiderivative of 3x squared is x cubed. And the antiderivative of negative four over x is minus four times ln absolute value of x. We could add a plus C to make it a general antiderivative, but we don't really need it. The fundamental theorem says that we can use any antiderivative, so we might as well use
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the simplest one, where c equals zero. Now we need to evaluate this antiderivative on the endpoints of negative one and negative five. And we usually write this as a vertical line with a negative one at the bottom and a negative five at the top to mean evaluation. In other words, the notation capital F of X between A and B means capital F of b minus capital F of A, which is what we need to compute for our antiderivative here. So now we just plug in negative five for x, and then we subtract
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what we get when we plug in negative one for x. In this example, you can see why it's important to write the antiderivative of one over x as ln absolute value of x not just ln of x, because ln of the absolute value of five, which is ln a five actually has an answer, whereas ln of negative five would not exist. I can simplify this expression a little bit, I get negative 125 minus four ln five minus negative one, plus four ln of one. Since ln
9:40:21
of one is zero, this becomes negative 124 minus four ln five. That's about negative 130 point 438. In this next example, we need to find the antiderivative for this expression y squared minus y plus one over the square root Why. Now we can't take the antiderivative separately of the numerator and the denominator, because that's just not how the quotient rule works for differentiation. So it doesn't work that way for anti differentiation either.
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Instead, let's try to simplify this expression to make it look more like something we can take the antiderivative of. So I'm going to rewrite the denominator as y to the one half power. And dividing by y to the one half is the same thing as multiplying by y to the negative one half distributed, distributing and adding exponents, I get y to the three halves minus y to the one half plus y to the negative one half. Now that's something I
9:40:31
can take the antiderivative of just using the power rule and reverse y to the three halves, becomes y to the five halves by adding one to the exponent, I divide by the new exponent. Now here, I get y to the three halves divided by three halves, and here are negative one half plus one is positive one half, I need to evaluate this between four and one. let me simplify a little bit. And now I'll substitute in values. Now four to the five halves is
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the same thing as four to the one half raised to the fifth power. So that's two to the fifth power, or 32. Similarly, for the three halves is four to the one half cubed, so that's two cubed, or eight, and four to the one half is just two. And one to any power is just one. And after some arithmetic, I get an answer of 146 15th. The Fundamental Theorem of Calculus, part two can be stated this way, for a function capital F with continuous derivative, the integral of the derivative is equal to the
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original function evaluated on the bounds of integration. In this video, I'll prove both parts of the fundamental theorem of calculus. The first part of the fundamental theorem of calculus says that if f of x is a continuous function, then the function g of x defined as the integral from a constant A to the variable x of f of t dt is differentiable, and has derivative equal to the original function, f of x.
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To prove this theorem, let's start with the limit definition of derivative. The derivative, g prime of x, by definition, is the limit as h goes to zero of g of x plus h minus g of x over h. Now g of x is defined as an integral from a to x. So g of x plus h is going to be the integral from a to x plus h, just plugging in x plus h for x of f of t dt. g of x is the integral from a to x of f of t dt. By properties of integrals. The integral from a to x plus h minus the integral from a to
9:40:55
x is just the integral from x to x plus h. Now informally, the integral from x to x plus h can be closely approximated by a skinny rectangle with height, f of x and width, H. And so this limit is approximately the limit as h goes to zero of f of x times h over h, which is just f of x. But let's make this argument a little more precise. Let's let capital M be the maximum value that f of x achieves on this little sub interval, and
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lowercase m be the minimum value achieved. In this picture, they occur on the endpoints of the interval from x to x plus h, but they could also occur somewhere in the interior. But we know that f of x does have to have a minimum value and a maximum value, since it's a continuous function by assumption on a closed interval. Now we know that the integral of f of t dt from x to x plus h has to be less than or equal to capital M times h and
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bigger than a to lowercase m times h. This is one of the properties of integrals, and can be verified visually by comparing this shaded red area to the small blue rectangle, which has area lowercase m times h, and comparing it to the area of the big rectangle, which has area capital M times h equivalently, the integral from x to x plus H of F of t dt divided by h has to be less than or equal to capital M and greater than or equal to little m. But
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the intermediate value theorem, which holds for all continuous functions, says that this intermediate value that lies between the minimum and maximum value of f has to be achieved as f of c for some C in the interval. Therefore, I cannot replace this integral in the limit expression above by just simply the value f of c for some c between x and x plus h. The value of C here depends on x and h. But as h goes to zero, C has to get closer and
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closer to x. And since f is continuous, this means that this limit is equal to f of x. We've now proved the first part of the fundamental theorem of calculus, that the derivative of g exists and equals f of x. The second part of the fundamental theorem of calculus says that if f is continuous, then the integral from a to b of f of x dx is equal to the antiderivative of lowercase F, which I'll denote by capital F, evaluated at B minus that antiderivative
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evaluated today. Part Two of the fundamental theorem follows directly from part one. Let's let g of x be defined as the integral from a to x of f of t dt. Then part one of the fundamental theorem of calculus tells us that g prime of x exists and equals lowercase f of x. In other words, capital G is an antiderivative for a lowercase F. Now g of the minus g of A is by definition, the integral from a to b of f of t dt minus the integral of a from
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a to a of f of t dt. The second integral is zero, since the bounds of integration are identical. So part two of the fundamental theorem of calculus is true if I use the antiderivative capital G. But the theorem is supposed to be true for any antiderivative. So let's let capital F be any antiderivative of lowercase F, we know that capital F of x has to equal capital G of X plus some constant since any two antiderivative for the same function differ by a constant,
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and therefore, capital F of b minus capital F of A is going to equal capital G of B plus C minus capital G of A plus C. The constant C cancels out, and we just get capital G of b minus capital G of A, which we already saw was equal to the integral from a to b of lowercase f of t dt. So the left side of this equation is equal to the right side. And the fundamental theorem of calculus Part two is proved for any antiderivative. This completes the proof
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of the fundamental theorem of calculus. This video is about the substitution method for evaluating integrals, also known as u substitution. As the first example, let's try to integrate to x sine of x squared dx. Now sine of x squared is the composition of the function sine and the function x squared. And notice that the function x squared has derivative to x, which is sitting right here and the integrand. I'm going to make the substitution u equals x
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squared, and then I'll write d u is equal to 2x dx. That's differential notation. To find d u, I take the derivative have X squared and then multiply by the differential dx, I can then rewrite the integrand as sine of u. And the 2x. dx becomes do after making this substitution, I can integrate, because the antiderivative of sine of u is negative cosine of u. And I'll add on the constant of integration. I'm not finished yet, my original
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problem was in terms of x, and now I've got a function in terms of u. So let's substitute back in since u is equal to x squared, I can replace that, and I have my final answer. To verify that this final answer is correct, that it really is the antiderivative of what we started with, let's take the derivative of our answer and make sure we get back the function to x sine of x squared. If we take the derivative of negative cosine of x squared plus C, then we get the derivative
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of a constant is zero, so we have the derivative of negative cosine, that's equal to sine of the inside function x squared times the derivative of the inside function using the chain rule. And we do in fact, get back to the integrand that we started with. Notice that we use the chain rule when taking the derivative to check our answer. Let's try some more examples of use substitution. When looking for what to substitute as you, it's good to look for a
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chunk that's in the integrand, whose derivative is also in the integrand. It's also good enough to just have a constant multiple of the derivative in the integrand. So in the first example, we might use the chunk one plus 3x squared, as are you. The derivative of that expression is six times x. And even though six times x isn't completely in the integrand, we do have a factor of x in the numerator, that's just a constant multiple away from the derivative
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of 6x. So let's write out d u, that's going to be 6x dx. And I'm going to go ahead and rewrite this as x dx is equal to one six d U. writing it this way, it makes it easy to substitute one six d u for x dx. And then in my denominator, my one plus 3x squared becomes u. I can rewrite this as one six times the integral of one over u d u you and I recognize that the antiderivative of one over u is ln absolute value of u. Substituting back in
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for you, I get a final answer of one six ln absolute value of one plus 3x squared, plus say, the absolute value signs are not really necessary in this example, since one plus 3x squared is always positive. As our next example, let's look at the integral of e to the 7x dx. one chunk with us here is u equals 7x. If we do that, then d u is just seven dx. And so we have dx is equal to 1/7 do substituting in we have the integral of e to the u times
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1/7 d u, I can pull the 1/7 out and integrate e to the u to the app, just either the U and substituting in for back for 7x, I get e to the 7x plus C. I encourage you to pause the video to check that these two answers are correct. By taking derivatives. You'll notice that you use the chain rule each time. Next, let's do an example with a definite integral, the integral from E to E squared of ln x over x dx. If we set u equal to ln x, then d u
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is the derivative of ln x, that's one of our x times dx. This is a much better choice of you than say setting u equal to x from the denominator, because then d u would just be dx. And when we did the substitution, nothing would really change. For definite integrals, we need to deal with the bounds of integration here e and d squared. We really have two options, worry about them now or worry about them later. I'll show you the worry about them. Now method
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first. Our bounds of integration E and E squared are values of Bax as we convert everything in our integral from x to you, we need to convert the bounds of integration from values of x to values of u also. Now, when x is equal to e, u is equal to ln IV, which is one, just using this equation. Similarly, when x is equal to E squared, u is equal to ln t squared, which is two. So as I rewrite my integral, I'm going to replace the bounds with one and
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two. And now my ln x becomes my u and my dx divided by x becomes my do, they're going to grow of UD u is equal to use squared over two, and I evaluate this between the bounds of u equals two and u equals one to get two squared over two minus one squared over two, which is one half. Notice that when we did the problem this way, we never actually had to get back to our variable x, we stayed in the variable u and evaluated. The second way
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of dealing with the bounds of integration is to worry about them later. Let's go back to the beginning of the problem. We're just about to substitute u equals ln x and d u equals one over x dx. Instead of substituting in for the bounds of integration, I'm going to temporarily ignore them and just evaluate the indefinite integral ln x over x dx, which I can substitute in as you times do, we can evaluate that to get you squared over two.
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Normally we'd have a plus c constant. But since we're ultimately going to be doing a definite integral, we don't really need the constant here. Now, just like when we're doing indefinite integrals, I'm going to get back to the variable x by substituting back in for you U is ln of x. So I square that and divided by two, and then I can go back to my original bounds of integration, those bounds are the x values of E squared, and E. Plugging
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in those bounds, I get ln of E squared quantity squared over two minus ln of E squared over two, which evaluates to two squared over two minus one over two, which is again, one half. This video gave some examples of use substitution to evaluate integrals. This method works great in examples like this one, where there's a chunk that you can call you whose derivative or at least a constant multiple of its derivative is also in the integrand.
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You've already seen how u substitution works in practice. In this video, I'll try to explain why it works. u substitution is based on the chain rule. Recall the chain rule says if we take the derivative of a function, capital F of lowercase g of x with respect to x, we get the derivative of capital F evaluated on the inside function g of x times the derivative of g of x. If we just write that equation in the opposite order, we have that an expression of the form
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f prime of g of x times g prime of x can all be wrapped up as the derivative of a composite function, f of g of x. Now if I take the integral of both sides of this equation with respect to x, on the right side, I'm taking the integral of a derivative. Well, the integral or antiderivative of a derivative is just the original function, capital f of g of x plus C. Now when we do use substitution, we're really just writing
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this equation down. We are seeing an expression of the form f prime of g of x times g prime of x dx, we're recognizing you as g of x and d u as g prime of x dx. So we're rewriting this expression as the integral of capital F prime of u d u do and that integrates to just capital F of u plus C. And then we're substituting back in for you To get capital f of g of x plus C, the beginning and end of this process are exactly the same as the
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left side and right side of our chain rule expression above. So when you're doing u substitution to integrate, you can thank this chain rule that's behind it all. This video introduces the idea of an average value of a function. To take the average of a finite list of numbers, we just add the numbers up and divide by n, the number of numbers. In summation notation, we write the sum from i equals one to n of Q i all divided by n. But defining the average
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value of a continuous function is a little different. Because a function can take on infinitely many values on an interval from a to b, we could estimate the average value of the function by sampling it at a finite Li many evenly spaced x values. I'll call them x one through x n. And let's assume that they're spaced a distance of delta x apart, then the average value of f at these sample points is just the sum of the values of f
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divided by n, the number of values are in summation notation, the sum from i equals one to n of f of x i all divided by n. This is an approximate average value of f, since we're just using n sample points. But the approximation gets better as the number of sample points n gets bigger and bigger. So we could define the average as the limit as n goes to infinity of the sample average. I'd like to make this look more like a Riemann
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sum. So I need to get delta x in there. So I'm just going to multiply the top and the bottom by delta x. And notice that n times delta x is just the length of the interval b minus a. Now as the number of sample points goes to infinity, delta x, the distance between them goes to zero. So I can rewrite my limit as the limit as delta x goes to zero, of the sum of FX II times delta x divided by b minus a. Now the limit of this Riemann sum in the
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numerator is just the integral from a to b of f of x dx. And so the average value of the function is given by the integral on the interval from a to b divided by the length of the interval. Notice the similarity between the formula for the average value of a function and the formula for the average value of a list of numbers, the integral for the function corresponds to the summation sign for the list of numbers. And the length of the interval
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B minus A for the function corresponds to n, the number of numbers in the list of numbers. Now let's work an example. For the function g of x equals one over one minus 5x. On the interval from two to five, we know that the average value of G is given by the integral from two to five of one over one minus 5x dx divided by the length of that interval, I'm going to use use of the tuition to integrate, so I'm going to set u equal to one minus 5x.
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So d u is negative five dx. In other words, dx is negative 1/5 times do. Looking at my bounds of integration, when x is equal to two, u is equal to one minus five times two, which is negative nine. And when x is equal to five, u is equal to negative 24. substituting into my integral, I get the integral from negative nine to negative 24 of one over u times negative 1/5. Do and that's divided by three. Now dividing by three
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is same as multiplying by 1/3. And as I integrate, I'm going to pull the negative 1/5 out and then take the integral of one over u, that's ln of the absolute value of u evaluated in between negative 24 and negative nine. The absolute value signs are important here because they prevent me from trying to take the natural log of negative numbers to evaluate get negative 1/15 times ln of 24 minus ln of nine, I can use my log rolls to simplify
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and get negative 1/15 ln of 24 over nine, that's negative 1/15 ln of eight thirds, and as a decimal, that's approximately negative 0.0654. So I found the average value of G. Now my next question is, does g ever achieve that average value, in other words, is there a number c in the interval from two to five for which GFC equals its average value? Well, one way to find out is just to set GFC equal to G's average value. In other words, set
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one over one minus five c equal to negative 1/15 ln of eight thirds, and try to solve for C. There are lots of ways to solve this equation. But I'm going to take the reciprocal of both sides, subtract one from both sides and divide by negative five. This simplifies to three over ln of eight thirds, plus 1/5, which is approximately 3.25. And that x value does lie inside the interval from two to five. So we've demonstrated that g does achieve
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its average value over the interval. But in fact, we could have predicted this to be true. Gs average value has to lie somewhere between GS minimum value and maximum value on this interval. And since G is continuous on the interval from two to five, it has to achieve every value that lies in between as minimum and maximum, including its average value. The same argument shows that for any continuous function, the function must achieve its average
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value on an interval. And this is known as the mean value theorem for integrals. Namely, for any continuous function f of x on an interval from a to b, there has to be at least one number c, between A and B, such that f of c equals its average value, or, in symbols, f of c equals the integral from A b of f of x dx divided by b minus a. This video gave the definition of an average value of a function, and stated the mean value theorem for integrals.
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If we rewrite the formula for average value a little, then we can see a geometric interpretation for average value, the area of the box with height the average value is the same as the area under the curve. This video gives two proofs of the mean value theorem for integrals. the mean value theorem for integrals says the for continuous function f of x, defined on an interval from a to b, there's some number c between A and B, such that f of c is equal
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to the average value of f. The first proof that I'm going to give us is the intermediate value theorem. Recall that the intermediate value theorem says that if we have a continuous function f, defined on an interval, which I'll call x 1x, two, if we have some number l in between f of x one and f of x two, then f has to achieve the value out somewhere between x one and x two. Keeping in mind the intermediate value theorem, let's turn our attention back
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to the mean value theorem for integrals. Now it's possible that our function f of x might be constant on the interval from a to b. But if that's true, then our mean value theorem for integrals holds easily, because f AV is just equal to that constant, which is equal to f OC for any c between A and B. So let's assume that f is not constant. Well, like continuous function on a closed interval has to have a minimum value and a maximum value,
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which I'll call little m, and big M. Now we know that F's average value on the interval has to be between its maximum value and its minimum value. If you don't believe this, consider the fact that all of us values on the interval have to lie between big M and little m. And if we integrate this inequality We get little m times b minus a is less than or equal to the integral of f is less than or equal to big M times b minus a. Notice that the first and the last integrals, we're
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just integrating a constant. Now if I divide all three sides by b minus a, I can see that little m is less than or equal to the average value of f is less than or equal to big M as I wanted. Now, I just need to apply the intermediate value theorem, with F average as my number L and little m and big M as my values of f of x one and f of x two. The intermediate value theorem says that F average is achieved by f of c for some C in between my x one and
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x two. And therefore, for some C in my interval a b. And that proves the mean value theorem for integrals. Now I'm going to give a second proof for the mean value theorem for integrals. And this time, it's going to be as a corollary to the regular mean value theorem for functions. Recall that the mean value theorem for functions, says that if g of x is continuous on a closed interval, and differentiable on the interior of that interval, then there's some number c in the
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interval, such that the derivative of g at C is equal to the average rate of change of G, across the whole interval from a to b. Let's keep the mean value theorem for functions in mind, and turn our attention back to the mean value theorem for integrals. I'm going to define a function g of x to be the integral from a to x of f of t dt, where F is the function given to us in the statement of the mean value theorem for integrals. Notice that g of A
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is just the integral from a to a, which is zero, while g of B is the integral from a to b of our function. Now, by the fundamental theorem of calculus, our function g of x is continuous and differentiable on the interval a, b, and g prime of x is equal to f of x. And by the mean value theorem for functions, we know that g prime of c has to equal g of b minus g of a over b minus a, for some numbers, C and the interval a b, if we substitute in
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the three facts above, into our equation below, we get f of c is equal to the integral from a to b of f of t dt minus zero over b minus a, which is exactly the conclusion that we wanted to reach. This shows that the mean value theorem for integrals really is the mean value theorem for functions where our function is an integral. And this completes the second proof of the main value theorem for integrals. So now I've proved the mean
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value theorem for integrals in two different ways. And I've used a lot of the great theorems of calculus along the way.
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