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0:00
Give me 50 minutes and I will make differential calculus click for you forever. You've heard the terms, right? Derivatives, differentiation. Maybe you've even seen dy/dx written down somewhere. They are the language of change. And honestly, they're non-negotiable for success in well, in a lot of fields. Physics, engineering, computer science, economics, statistics, and of course, deep learning. At the very heart of differential calculus, you find this. Okay, look at this table.
0:44
This is the toolbox you will master. It probably looks a little intimidating right now. I get it. We've got categories like basic rules, essential functions, trigonometric functions, and then these combination rules. Down at the bottom, you can see a function f ofx. And right next to it, its derivative frime ofx. This table is the core of differential calculus. It looks like a random list of rules to memorize. But here's the thing, it's not. By the end of this video,
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you won't have memorized this table. That's not the goal. No, you will have built it from a single idea. You will understand precisely why the derivative of a function, let's say f ofx= x^2, is exactly 2x. And from that one core insight, you will see how every other rule logically follows. From the master key, which is the chain rule, to the truly exotic stuff, like the derivative of ark
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sine of x becoming 1 / the square<unk> of 1 - x^2. No magic, just mechanics. Let's get started. Let's talk about a car's journey. A car's position in meters, we'll call it P, after t seconds, is given by a really simple function. P of t= t^2. So at t = 1, it's at 1^2, which is 1 m. At t = 2 seconds, it's at 2^2. That's 4 m. And at t = 3 seconds, it's at 3^ 2, 9 m. Simple enough, right?
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But here's the question. What is the car's exact speed at the single instant t= 2 seconds? And this leads to a huge problem. I mean, think about it. How do you measure speed over zero time? The formula for speed is distance divided by time. If the time interval is zero, you're dividing by zero. And you just you can't do that. This right here is the central problem of differential calculus and the tool we invented to solve it is called the limit. The limit is like
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the microscope of calculus. It lets us zoom in on a single instant by looking at what happens as we get infinitely close to it. So let's try that. We can't calculate the instantaneous speed, but we can calculate the average speed. And the formula for average speed is just the change in position divided by the change in time or as you might see it delta P over deltat t. So let's do a quick calculation for the average speed. Our context is still p of t= t ^2. Let's find the
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average speed between t = 2 and t = 3. First, the change in position. That's p of 3 - p of 2, which is 3^ 2 - 2^ 2. That's 5 m. Next, the change in time. That's just 3 - 2, which is 1 second. So, the average speed is 5 / 1, which gives us 5 m/ second. But that's a really crude average, isn't it? It's over a whole second. To get closer to the speed at = 2, we need to squeeze the interval.
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Now, watch what happens as we get closer to t= 2. Look at that first row. The time interval is from 2 to 3. As we just saw, the change in time delta t is 1. The change in position delta p is 9 - 4, which is 5. And the average speed is 5 m/ second. Okay, got it. Now, let's squeeze. Let's look at the interval from 2 to 2.1. Now the change in time is just.1. The change in position is 4.41 - 4 which is41. So the average speed is41 /.1 which is 4.1 m/s. Interesting. Let's
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get even closer. The interval from 2 to 2.01. Delta T is now tiny just 001. Delta P is 4.0141 minus 4 which is 041 and the average speed is 4.01 m/s. Do you see it? The pattern is inescapable. The number is being squeezed towards a single value. What value is it approaching? Four. So we can make a statement. The limit of the average speed as the time interval delta t approaches zero is four. Now this isn't just some abstract algebraic idea. It's a geometric one too.
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First think about the secant line. The average speed between two points on our curve is just the slope of the line connecting them. Second, as we shrink the interval, as we slide that second point closer and closer to the first one, watch that secant line. It begins to pivot. And third, as it gets infinitely close, it settles into one final perfect position. It becomes the tangent line. The one line that just skims the curve at that single point. Its slope is the instantaneous
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speed. Okay. Now let's formalize all of this with the proper notation. The limit of the function f ofx as x approaches c equals l. Let's connect this notation directly to our car example. Look at this table. First component li or limb. That just means we're doing the approaching process. Next we have x approaches c. In our example, what was approaching what? Our time interval deltat t was approaching zero. So for us it's deltat t approaches zero. Then you have f ofx. That's the
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function we are observing. For us, that was our average speed formula. P of 2 + delta t minus p of 2 all divided by delta t. And finally, L, the limit, the value our function squeezed towards. And what was that value? Four. So, putting it all together, the precise mathematical question we just answered is this. The limit as delta t approaches zero of the function 2 + delta t^ 2 minus 2^ 2 all divided by delta t equals 4. We solved this by observing a pattern in a
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table. But that's not always practical. We need a way to solve this algebraically. We're going to generalize that exact process into a single powerful formula. This formula will take any function f ofx and forge a new function frime ofx that gives the instantaneous rate of change at every single point. This is the master equation of differential calculus. frime of x equals the limit as h approaches zero of fx + h minus fx all / h. Let's deconstruct the formula. Check out this
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table. It's not just a random string of symbols. It's a story. First component in our formula, H. What does it mean? It's just a tiny change in the input. For our car example, this was just delta t, a small change in time. Simple enough. Next up, f ofx plus h minus f ofx. This is the change in the output. the rise in our car example that was P of T plus H minus P of T just the change in position.
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Okay. Now let's look at the whole fraction. The rise divided by H. The run it's the average rate of change rise over run for our car. This was the average speed over that tiny interval h. And the most important part, limb, the limit as h approaches zero. What's that? It's the squeeze. It's the process of finding the instantaneous speed from the average speed. And finally, frime of x. That's the result. The new function for the instantaneous rate of change for our car.
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This becomes v of t, the new velocity function. See, it's not a random string of symbols at all. It's a story. The story we just uncovered. Now for our first boss battle, we are going to derive the derivative of f(x)= x^2 using the limit definition from first principles. Step one, set up the formula. We just plug our function in. frime of x equals the limit as h approaches zero of x + h^2 minus x^2 all divided by h. Step two, expand the numerator. We just expand that x + h^2
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term. So now we have the limit as h approaches zero of x^2 + 2 xh + h^2 - x^2 all divided by h. Step three simplify the numerator. Now watch this. This is the cool part. We have an x^2 at the beginning and a minus x^2 at the end. They cancel each other out. This leaves us with the limit as h approaches zero of 2xh + h^2 / h and cancel. This is the magic step. The whole reason we can't just
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plug in 0 for h is because we'd be dividing by 0. But now look, we can factor an h out of the top, giving us h * parentheses 2x + h / h. The h on top cancels the h on the bottom. The problem is solved. We're just left with the limit as h approaches zero of 2x + h. Which brings us to our final move. Step five. Evaluate the limit. Now that the denominator is gone, it's finally safe to let h become zero. So we have 2x + 0, which is just frime of x = 2x. And we're done. We
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have rigorously proven from first principles that if f ofx= x^2, its derivative is frime ofx= 2x.
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So let's apply this back to our car. The car's position function was p of t= t ^2. Its velocity function the derivative is therefore v of t = p prime of t which equals 2 t. The original question was what is the speed at t = 2? v of 2 = 2 * 2 which is 4 m/s.
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It's the exact same number we found by squeezing the intervals. The beast is tamed. We now have a formal algebraic machine that confirms our intuition perfectly. So, in this section, we're going to build our foundational library of derivatives. Think of this as our spell book. This is where intuition beats memorization every single time. Let's start with the absolute basics. First up, the derivative of a constant. That's a function like f ofx= c. The formal proof is right
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here. We take the limit as h approaches zero of fx + h - fx all over h. Well, f of anything is just c. So this becomes the limit of c minus c / h, which is the limit of 0 / h, which is just zero. The intuition, a horizontal line has zero slope everywhere. Simple as that. Second, the derivative of a line. Let's say f ofx= mx. Again, we use the limit definition. We get the limit as h approaches
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0 of m * the quantity x + h - mx all over h. If we distribute that m, we get mx + mh - mx all over h. The mx terms cancel out. We're left with mh over h. The h's cancel and the limit is just m. And the intuition makes perfect sense. A line has a constant slope m everywhere. Now for something special, the number e, which is about 2.718, is the single most important base in all of calculus.
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The function f ofx= e tox is its own derivative. Let that sink in. the derivative with respect to x of e to x i e to x. This means for the graph of y = e to x, the slope of the tangent line at any point x is exactly equal to the value of the function y at that same point. Let's trace along the curve. We'll check a few points. First at x= 0 the value is y = e to 0 which is 1. The slope of the tangent line at that point is m = 1. They match. Okay, let's keep tracing. How about at x=
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1? The value is y = e 1 which is about 2.718. And the slope the slope m is also about 2.718. They match. One more. Let's go to x=1. The value y = e to the -1 which is about 0.367. And guess what the slope is? The slope m is about 0.367. They match. This property is the very definition of the number e. It is the fundamental language of growth, which is why it's everywhere in physics, finance,
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and biology. Next up, its inverse, the natural logarithm. f ofx= ln of x. Its derivative is beautifully simple. The derivative with respect to x of the natural log of x is 1 /x. Let's build the intuition for this one, too. We'll trace the graph of the natural logarithm of x. First, let's check a point close to zero, like x= 0.1. The value ln of 0.1 is about -2.3. But the slope the slope is
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1 / 0.1 which is 10. Very steep. Okay, let's move to x= 1. The value ln of 1 is exactly 0ero. The slope is 1 / 1 which is 1. A perfect unit slope. Let's go way out to x= 10. The value ln of 10 is about 2.3, but the slope is now 1 / 10, which is just 0.1, almost flat. The 1 /x function perfectly describes the slope of the natural log. Finally, the trigonometric functions sin of x and cosine of x. These two are a beautiful pair. The derivative of sin of x is cosine of x and
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the derivative of cosine of x is sin of x. Let's focus on that first one. How is cosine of x the slope of sin of x? Let's trace the sine wave. We start at x= 0. The value s of 0 is 0. But look at the graph. It's at its steepest point going up. The slope is exactly one. And what's cosine of 0? It's one. Steepest rise. Okay, let's slide over to the peak. At x=<unk> / 2, the value s of p<unk>
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/2 is 1. But at a peak, the graph is momentarily flat. The slope is zero. And cosine of p<unk> /2 is 0. It's a perfect match. Now down to x= pi. The value s of pi is zero again. But now the graph is falling at its steepest point. The slope is -1. And what's cosine of pi? -1. Steepest fall. One last stop at the bottom of the trough. X= 3<unk> / 2. The value s of 3<unk> / 2 is -1. But again, it's a trough, so it's momentarily flat. The slope is 0 and cosine of 3<unk> / 2 is zero.
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It works perfectly. We now have our core library, our essential spell book. These are our building blocks for everything that comes next. We've built our foundational library. We know the derivatives of x^2, e to x, and sin of x. That's great. But real world functions are messy. The position of an object isn't just t ^2. It's something like this. P of T = 4 T cubed - 5 T ^2 + 7 T + 10. Now imagine trying to plug that into the limit definition. You'd have the limit as h approaches
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0 of 4 * t + h cubed minus 5 * t + h 2 minus the original function all over h. Uh, the algebra would be an absolute nightmare. Hard mode. But I'm going to show you how to find the derivative of that entire function in 15 seconds. We're switching to easy mode. To do this, we need our basic toolkit. Just three rules. One, the power rule. Two, the constant multiple rule. and three the sum and difference rule. Let's start with rule one, the power rule. This is the workhorse
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of differentiation. The formula looks like this. d / dx of x to the n = n * x to the nus1. The algorithm is just two simple steps. First, bring the old exponent down in front as a multiplier. Second, subtract one from the old exponent to create the new exponent. That's it. Let's see the power rule in action. First up, our input function is x^2. That's already written as x ^ of n. So, we apply the rule n * nus1. The 2 comes down in front. So, we get 2 * x to the power of
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2 - 1. The result 2x. Simple. Next one, x to the 5th. Same process. The five comes down in front, giving us 5 * x to the 5 - 1 power. The result is 5x to 4th. You're getting the hang of it. Okay. What about just x? Well, remember that x is really x to the first power. So, the one comes down 1 * x to the 1 - 1 power. And x to the 0 power is just 1. So the derivative is 1. Now for a tricky one,
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the square root of x. First we have to rewrite it. The square root of x is the same as x^ the 1/2 power. Now we apply the rule. The 1/2 comes down * x to the 1/2 - 1 power. Our final result is 12x^ - 1/2 power. Last one. 1 /x^2. We can rewrite this using a negative exponent. That's x^ the -2. Now apply the rule. The -2 comes down in front time x^ -2 -1 power. That gives us -2x to the -3.
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It's just a mechanical process. All right. On to rule two, the constant multiple rule. What happens when you have a number in front like 5x cubed? The rule is d over dx of c * f ofx = c * frime ofx. The constant just comes along for the ride. Let's do a quick example. Find the derivative of f ofx= 5x cubed. Step one, identify the constant c= 5 and the function x cubed. Step two, differentiate the
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function part. The derivative of x cubed is 3x^2. Step three, multiply by the constant 5 * 3x^2. The result, 15x^2. See, the five just hung out until the end. Here's another one. Find the derivative of g ofx = -2 cosine of x. The constant is -2. The function is cosine of x. We know the derivative of cosine of x is sin of x. So we multiply -2 * sin of x. The two negatives cancel out and the
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result is 2 sin of x. And that brings us to rule number three, the sum and difference rule. Our final tool lets us break big problems into small easy ones. The formula is the derivative of f(x) plus or minus g ofx equals frime ofx plus or minus g prime ofx. It means you can differentiate each term one by one completely ignoring the others. For example, let's find the derivative of h ofx= x 4 - 7x. First just look at the first term. The derivative of x^ 4th power is 4x cubed. Now
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look at the second term. The derivative of -7x is just -7. Now combine them. 4x cub - 7. That's our answer. We just took it piece by piece. Okay, the challenge. Find the derivative p prime of t of our original monster function. P of T = 4 T cubed - 5 T ^2 + 7 T + 10. Ready? Go. First, we use the sum rule. This means we can do this term by term. We can find the derivative of 4t cubed, then subtract
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the derivative of 5t ^2, and so on. Step two, let's differentiate each piece. The derivative of 4t cubed becomes 4 * 3 t ^2 which is 12 t ^2. The derivative of 5t ^2 becomes 5 * 2t which is 10 t. The derivative of 7t becomes 7 * 1 which is just 7. And the derivative of the constant 10 is zero. So step three assemble the final answer. P prime of T equ= 12 T ^2 - 10 T + 7. And we're done.
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What would have been a nightmare of algebra with the limit definition, we just solved in seconds. This is the power of our toolkit. We have our basic toolkit. We've totally mastered addition and subtraction. But what about multiplication and division? What about the star and the slash? Let's say I give you this function h ofx = x^2 * sin of x. What's the derivative? It's so tempting to just do them separately, isn't it? But that's not how it works. We need a special rule. Rule
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number four, the product rule. This is the one you use when two functions are multiplied together. The formula is d / dx of f(x) * g ofx = frime ofx * g ofx plus f ofx * g prime of x. I know, I know it looks a little strange, but there's a rhythm to it. Here's how you actually say it. The derivative of the first time the second plus the first time the derivative of the second. Let's try
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it on our example. Our example was h ofx= x^2 sin of x. We're going to apply the product rule step by step. Step one, identify the two functions. The first function f ofx is just x squared. The second function g ofx is sin of x. Easy enough. Step two, find their individual derivatives. The derivative of x^2, that's frime ofx= 2x. We know that one. And the derivative of sin of x from our library,
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that's g prime of x= cosine of x. Okay, we have our four pieces. Now for step three, assemble the four pieces using the formula. Remember the rhythm derivative of the first time the second. So that's 2x * sin of x plus the first time the derivative of the second. So that's x^2 * cosine of x. There it is all laid out. Step four is just to simplify. H prime of x = 2x sin of x + x^2 cosine of x.
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That's it. That's our answer. Not so bad when you follow the pattern, right? Okay. What about division? This one's the big one. It's for finding the derivative of one function divided by another. Something like e to x / x cubed. This is rule number five, the quotient rule. And look at this formula. d over dx of f ofx over g ofx equals uh frime ofx g ofx minus f ofx g prime of x all over
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g of x²ared now nobody remembers that seriously it's just a mess but everyone who has ever learned calculus remembers the pneummonic here's how it works the function on top is high and the function on the bottom is low. So f ofx over g ofx becomes high over low. And the pneummonic, the little song you sing to yourself, goes like this. Low d high minus high d low over the square of what's below. Say it with me. Low d high minus high d low over the square of what's below.
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Let's break that down piece by piece. Low is just the bottom function g ofx. d high is the derivative of the top frime ofx. High is the top function f ofx. And d low is the derivative of the bottom g prime of x. Finally, the square of what's below is just g of x squared. See, it maps perfectly to that messy formula, but it's so much easier to remember. All right, let's use it. Example, h ofx= e x / x cubed. We're applying the quotient rule. Step one, identify high and
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low. The high function is e to x. The low function is x cubed. Step two, find their derivatives. d high and d low. The derivative of e to the x is just e to x. So d high is e to x. The derivative of x cubed is 3x^2. So d low is 3x^2. Step three, assemble the pieces using our pneummonic. Let's chant it. Low d high. That's x cubed * e to x minus high d low. That's e to x * 3x^2 over the square of what's below. That's x cubed all^ 2. Perfect. Now we just have to clean it up. Step
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four. Simplify. The numerator is x cub e x - e x * 3x^2 and the denominator x cub^ 2 becomes x 6. Okay, let's keep simplifying. We can factor out some common terms from the numerator. It looks like both terms have an e to x and an x^2 in them. Let's pull that out. That gives us e x * x^2 * the quantity x - 3 all over x 6. We're almost there. We can cancel x^2 from the top and bottom. The x^2
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on top goes away and the x 6 on the bottom becomes x to 4th. We did it. Now for the payoff. Remember back in our library section, I just gave you the derivative of tangent of x. I said it was secant^ squar of x. It felt like magic. It's not magic. We can now prove it. We know from trig that tangent of x is the same as sin of x / cosine of x. That's a quotient. We can use our rule. High is sin of x.
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So d high is cosine of x. Low is cosine of x. So d low is negative sin of x. Let's chant the rule one more time. Low d high minus high d low. Let's plug it in. The derivative of tangent of x equals low which is cosine of x* d high which is also cosine of x minus high which is sin of x * d low which is negative sin of x. All of that is over the square of what's below. So cosine of x^ squar okay let's simplify that numerator cosine * cosine is cosine^ 2 of x and s * negative sign
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is sin^ squar but we were subtracting it a minus and a minus make a plus so we get cossine^ 2 of x plus sin^ 2 x all over cossine^ 2 of x. Now look at that numerator. What is cosine^ 2 of x plus sin^ 2 of x? It's the most famous identity in all of trigonometry. It's 1. Our entire formula just became 1 / cossine^ 2 of x. This is amazing. And what is 1 / cosine of x? That's the definition of secant of x. So 1 / cossine^ 2 of x is just secant^ 2 of x. The derivative of tangent of x
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is secant^ 2 of x. We've done it. We just proved one of our library rules completely from scratch. See how it all connects? No magic, just rules building on rules. We can now handle addition, subtraction, multiplication, and division. But we have one final challenge. The boss battle. What do you do with a function inside of another function? Think about h ofx= sin of x^2 or even worse h ofx = x cubed + 5x all raised to the 7th power. The rules we have can't touch
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this. Trying to multiply out x cub + 5x 7 times would take hours. We need a new rule. This isn't just a new rule. This is the rule. This is the master key of differential calculus. It's called the chain rule. It is the single most powerful and common derivative rule you will ever use. Master this and you've mastered differentiation. The secret Russian nesting dolls. C functions as an outer function and an inner function. For sine of x^2, the outer function is sine of something.
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The inner function is x^2. For x cub + 5x to the 7th, the outer function is something to the 7th power and the inner function is x cubed + 5x. Now here's the chain rule formula. If h ofx= f of g ofx, its derivative is h prime of x= frime of g ofx* g prime of x. Okay, that looks a little abstract. I know. Here's what that means in plain English. First, differentiate the outer function, leaving the inner function completely untouched. Just leave it there. Second, then multiply your
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result by the derivative of the inner function. You see frime of g ofx * g prime of x. We have a little saying for this. It goes derivative of the outside time derivative of the inside. That's it. That's the rhythm you need to remember. Derivative of the outside times derivative of the inside. Let's use this master key to unlock these impossible looking problems instantly. Example one, the polomial power. Find the derivative of h ofx= x
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cub + 5x all to the 7th power. Okay, we're going to do this step by step. First identify outer and inner. The outer function, the shell, is something to the 7th power. Its derivative is 7 * something to the 6th power. The inner function, the core, is x cubed + 5x. Its derivative is 3x^2 + 5. Okay. Now, let's apply the chain rule. Step one, differentiate the outer keeping the inner. So our result is 7 * and we just slot that inner
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function right back in there. 7 * the quantity x cub + 5x all raised to the 6th power. Now step two multiply by the derivative of the inner. We found that the derivative of x cub + 5x is 3x^2 + 5. So the final answer is h prime of x = 7 * x cub + 5 x^ 6 power* 3x^2 + 5 done. No algebra nightmare. Just a clean two-step process. This is the power of the chain rule. Okay. Example two. The trigonometric function. Let's find the derivative of h ofx= sin of x^2. Step one,
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identify outer and inner. The outer function is s of something. Its derivative is cosine of something. The inner function is x^2 and its derivative is 2x. All right, step two. Let's apply the chain rule. Derivative of the outer keeping the inner is cosine of x^2. Now multiply by the derivative of the inner which is 2x. The final answer is h prime of x= cosine of x^2 * 2x. We usually write the polomial part first just for neatness. So the final answer is written as
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h prime of x = 2x * cosine of x^2. With the chain rule in our toolkit, we can differentiate almost any function you'll encounter. Our set of tools for yals f ofx is now complete. So far, we've been dealing with functions that are really well nicely behaved. We call this explicit. Y is explicitly defined in terms of x. It's pretty easy. You've got things like y = x^2 or maybe y = sin of x /
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e x. You can always get y on one side and all the x stuff on the other. But what about relationships where x and y are all tangled together where you can't easily get y by itself. Think about the equation of a circle. x^2 + y^2 = 25. This is an implicit relationship. If you try to solve for y, you get this mess. y = plus or minus the<unk> of 25 - x^2. It's messy. It's not one function, but
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two. It's a pain. This is where we learn to hack the system. What I'm about to show you isn't a new rule. Not at all. It's a clever application of the chain rule that lets us solve problems that look impossible. It's called implicit differentiation and it feels like a cheat code. Here is the entire trick. The one key idea you need to remember. We're going to take the derivative of equations with respect to x. First, when we see an x term like x^2, we differentiate it normally. The
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derivative of x^2 is just 2x. Simple. Second, when we see a y term, we have to remember that y is secretly a function of x. So the chain rule must apply. Let's look at this. The standard rule, the derivative with respect to x of x^2 is just 2x. Now for the chain rule application, what's the derivative with respect to x of y^2? Well, the outer function is something squared. So its derivative is 2 * something. The inner function is y. So its derivative is dydx. So the final
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answer is 2y * dydx. That's it. That is the entire hack. Every time you differentiate a term with y, you do it normally and then you must multiply by dy over dx because of the chain rule. Here's the algorithm. It's four simple steps. One, take the derivative of both sides of the equation with respect to x. Two, when you differentiate a term with y, remember to multiply by dy over dx. Three,
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you'll get an equation with x, y, and dydx all mixed up. Don't panic. Four, just use algebra to solve for dy dx. Let's do an example. Example one, the slope of a circle. We want to find the formula for the slope dydx of the circle x^2 + y^2 = 25. Step one, differentiate both sides with respect to x. So we take ddx of x^2 + y^2 and set it equal to ddx of 25. All right. Now for step two, let's
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go term by term. The derivative of x^2, that's 2x. Easy. The derivative of y^2. Ah, this is our key trick. It's 2y * dydx. and the derivative of 25 that's just zero. Our new equation is 2x + 2y * dy over dx equals 0. Okay, last part, step three. Now we just solve for dydx. We subtract 2x from both sides to get 2yd dy over dx = -2x. Then we just divide by 2y and we get dy by dx =x over y. This
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is our answer. But does it work? Let's not just trust it. Let's prove it. Let's test a point. How about the point 3a 4? According to our formula, d y / dx =x / y. So the slope at 3a 4 is -34s. Looking at the graph, yeah, that looks plausible. Okay, test point two. 0a 5. This is the very top of the circle. The tangent line should be completely flat. The slope should be zero. Our formula gives us dy / dx = 5, which is zero. It works perfectly. One more test point 3. 5,
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0. This is the far right side. The tangent there should be vertical. The slope should be undefined. Our formula gives us dydx = -5 / 0. Division by zero. It's undefined. The hack works. It gives us the correct slope at every single point on the circle. Are you ready for your victory lap? Let's use this clever trick to do something amazing. Back in section 3, I just gave you the derivative
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for arine of x. It was 1 over the<unk> of 1 - x^2. It felt like it came from nowhere. Like magic. We're about to kill the magic. We are going to derive it ourselves right now. The victory lap. Finding the derivative of arine of x. The problem is find the derivative of y= arine of x. Okay, first we rewrite it implicitly. If y equals arine of x, that's the same thing as saying sine of y equals x. Now this is something we can work with. Okay. Step two,
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differentiate both sides with respect to x. On the left side, the derivative of s of y becomes cosine of y * dy / dx. On the right side, the derivative of x is just 1. Our new equation is cosine of y * dy / dx = 1. Step three, solve for dy / dx. We just divide by cosine of y. We get dy / dx = 1 / cosine of y. We're so close. This is the right answer, but it's in terms of y. We need it in terms of x. Time for a quick trip to trigonometry. We know that sine of y equals
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x. Let's visualize this with a right triangle. Imagine an angle y. S is opposite over hypotenuse. So if s of y is x over 1, the opposite side is x and the hypotenuse is 1. Using the Pythagorean theorem, the adjacent side must be the square roo<unk> of 1^2 - x^2. So what's cosine of y? It's adjacent over hypotenuse. That means cosine of y equals the<unk> of 1 - x^2. Now for the final step, we substitute our finding for cosine of y. Our derivative was dy / dx = 1 / cosine of
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y. Our triangle just showed us that cosine of y is the<unk> of 1 - x^2. So we get dy by dx = 1 / the square<unk> of 1 - x^2. And there it is. We have just proven the derivative of ar sine of x. It's not a magic formula you need to memorize. It's a direct logical consequence of the chain rule. a consequence that you now have the power to derive yourself. We have built the entire machine from the limit to the basic rules to the master key of the chain rule to this clever
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hack of implicit differentiation. Our toolbox is complete. Look at this. Your completed toolbox. from one simple question about a car's speed. We built the entire machine of differential calculus. You didn't just memorize formulas. You built every single rule from the ground up. the basic rules, the essential functions, the trigonometric derivatives, the product rule, the quotient rule, the master key chain rule, even the clever hack of implicit differentiation. You now possess
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the complete toolkit for differentiation. The promise from the beginning fulfilled. So what's next? Integral calculus. the other half of the story. You've built the machine. Now go use it.
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