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Top 10 Derivative Rules · Acedra
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Top 10 Derivative Rules
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0:00
for some function f ofx its derivative fime of x is an equation that gives the instantaneous rate of change of this function at any single point and that instantaneous rate of change can be represented by the slope of a line tangent to this function at that point and from Newton's quotient we know that the derivative which gives us the slope of the tangent line at any point is just equal to the limit of the secant slopes which would just be the difference in the Y
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values divided by the difference in the X values of those points where the difference in these X values which we call H is approaching zero and in this video I want to show you the top 10 derivative rules that you can use to find the equation of a derivative of a function without having to use Newton's quotient must know rule number one the power rule of derivatives for this rule if you want to find the derivative with respect to X for a power function X
1:00
to the^ of n notice the base of the power is our variable X and the exponent n can be any real number we can find its derivative using the power rule which says you can take the exponent and write it as the coefficient of the power you keep the base of the power the same so it remains an X and then you take the exponent and reduce it by one so the new exponent becomes n minus1 and now let me show you a few examples where we can apply this rule looking at the first example
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I see that I'm finding the derivative of a power function based on the power rule of derivatives I know I'm allowed to take the exponent five and write it as the coefficient of the power I keep the base of the power as X and then reduce the exponent by 1 and 5 - 1 is 4 so my derivative of x 5 is 5X ^ 4 applying that same rule to example number two where we're differentiating this power function x^ 3 4 I can take the exponent of 34 and write it as the coefficient of my power
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keep the base of the power as X and then reduce the exponent by 1 3/4 - 1 that's 3/4 - 4/4 which is - one4 and then a proper Final Answer shouldn't have a negative exponent so I can take this power which has a negative exponent and I can move it down into the denominator with that four to change the sign of the exponent and then lastly for example three I know in finding the derivative I can take the exponent and write it as the coefficient keep the base of the power as
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X and then reduce the exponent by 1 --3 - 1 is -4 and then since this power has a negative exponent I could write its reciprocal so that it has a positive exponent making my final answer become -3 over x ^ of pos4 must know derivative rule number two the constant rule this is perhaps the easiest of all derivative rules if you're finding the derivative with respect to X of any constant so we'll say the derivative of C where C stands for any constant the derivative of any constant
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is just zero so for example if I asked you to differentiate the constant five the derivative of five is just zero and there's a couple ways to understand that remember a derivative is the instantaneous rate of change of a function well how does a constant value change it it doesn't that's what this answer is representing and then graphically speaking if I were to graph this function 5 what does yal 5 look like yal 5 would just be a horizontal line through five on the Y
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AIS and since a derivative is the instantaneous rate of change or slope of a function well what is the slope of this horizontal line anywhere well if I were to pick any two points on this function and calculate the slope between those two points remember slope is rise over run and since there is no rise between those two points I know my slope would just be zero know derivative rule number
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three the constant multiple rule for this rule if you're finding the derivative of a constant times a function of X what that's equal to you can just take that constant factor and multiply that by the derivative of the function of X this would equal C multipli the derivative of the function of X and let me give you two examples of how this works looking at our first example I'm finding
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the derivative of this function and that function is made made up of a constant Factor 3 multiplied by a function of x x ^ 5 so what you're allowed to do using the constant multiple rule is you can take that constant factor of three and put it in front of the derivative calculation and then you know how to find the derivative of x 5 just by using the power rule the derivative of x the^ of 5 is just 5 x 4 and then you could simplify these constant factors of 3 * 5 you
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could simplify that to 15 making the derivative 15 x to the^ of 4 and you'll probably notice a shortcut that works if we have power functions like this you can just take the exponent and multiply it by the coefficient to give you 15 and then reduce this exponent by one to give you four let's try using the constant multiple rule for example two as well we're finding the derivative with respect to X of this function which is 8 * X it's a product of a constant
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and a function of X so this constant factor 8 can be moved out front and if it helps this X has an invisible exponent of one so in calculating the derivative I can leave that constant factor of 8 and multiply that by the derivative of x using the power rule I take the exponent and write it as the coefficient keep the base and then reduce the exponent by one making the exponent zero and X to the^ of 0 means there are no factors of X we could just
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erase it which means the overall derivative is equal to just 8 * 1 which is of course 8 must know derivative rule number four the sum and difference rules of derivatives for these rules if you're finding the derivative of a sum or difference of two functions of X so f ofx plus or minus G of X what you're allowed to do when finding the derivative of a sum or difference of functions is just set that equal to the sum of the derivatives of each of those functions
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separately so it would equal to the derivative of FX plus or minus the derivative of G of X and now let me show you with an example how this works in this example I'm finding the derivative with respect to X of this polinomial function which is made up of three different terms to find its derivative I'm allowed to just differentiate all three of these terms separately and then just add and subtract their derivatives and now we can just differentiate all three of those terms
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the derivative of 5x to ^ of 6 using the power rule I know I would multiply this 6 by this 5 to get 30 keep the base of the power as X and then reduce the exponent by 1 making it five minus the derivative of 4X Cub 3 * 4 is 12 keep the base of the power as X and reduce the exponent by 1 making it two and then my last term the derivative of any constant is zero but you wouldn't have to write plus 0 making the derivative of this polinomial function equal to 30X 5 - 12x^2 must
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know derivative rule number five the product rule for this rule if you're finding the derivative of a product of two functions of X so if you have the derivative of FX being multiplied by G of X the product rule would say that the derivative of this product is equal to the derivative of the first function of X in the product so frime of X multiplied by the second function of x g of X
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plus the derivative of the second function of x g Prime of X multiplied by the first function of x f x so if we think of FX as the first function and G ofx as the second function product rule says when finding the derivative of the product of those functions you do the derivative of the first * the second plus the derivative of the second * the 1 and now let me show you with an example how we
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can apply this rule in this example we're asked to find the derivative with respect to X of a product of two functions of X for that reason we need the product rule of derivatives the product rule of derivatives tells me that I would have to find the derivative of the first function of X and the derivative of 3x - 5 would just be equal to 3 that needs to be multiplied by the second function
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of X that's the first part of the formula done now I have to add the derivative of the second function times the first so the derivative of x^2 + 1 is 2 X and that needs to be multiplied by the first function of X in the product which is 3x - 5 and now this is the derivative of this product but we could simplify it into standard form so I could expand the three into that binomial and expand the
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2x into that binomial and then collecting my like terms 3x^2 + 6 x^2 is 9 x^2 I have my derivative of 9 x^2 - 10 x + 3 must no derivative rule number six the quotient rule in this rule if you want to find the derivative of a quotient of two functions of X let's say we want the derivative of FX / G of X I'll refer to f x as the top function and G of X as the bottom function when differentiating that
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quotient that's equal to the derivative of the top function multiplied by the bottom function minus the derivative of the bottom function multiplied by the top function and that all needs to be divided by the bottom function G ofx squared so let me show you with an example how you can use this rule in this example notice that we're finding the derivative of a quotient of two functions of X for that reason you need to use the quotient Rule and the quotient rule states do the
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derivative of the top function the derivative of 3x - 4 is 3 and that gets multiplied by the entire bottom function x^2 + 5 minus the derivative of the bottom function the derivative of x^2 + 5 is 2X and that gets multiplied by the top function 3x - 4 and then all of that needs to be divided by the bottom function squared so it needs to be divided by x^2 + 5 squar and then of course you need to be able to simplify your derivative equation so the numerator needs to be expanded
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and simplified I would distribute the three to both terms in that binomial and distribute the -2X to both terms in that binomial and then in the numerator I have a couple like terms I have 3x^2 - 6x^2 which I could collect together to make - 3x^2 so I'll do that below and this is the final simplified version of the derivative of this quotient must know derivative rule number seven the chain rule this rule will allow you to find the derivative of composite functions meaning if
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you have a function of a function so that could be F of G of x f is our outer function and G is is our inner function so we have a function of a function when that happens when you have a composite function you can find its derivative by differentiating the outside function with respect to the inside function meaning find the derivative of f so F Prime but then leave the inside function exactly as it is but then that has to be multiplied by the derivative of the inside
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function so we would multiply this by G Prime of X and this chain rule can be applied app to all different types of composite functions since we've mostly been working with power functions up till now I'll show you how this rule gets applied to a power function if you're finding the derivative of a power of a function meaning you have a power function but the base is a function of X not just
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X so let's say we have G of x to the^ of n we could differentiate that by following the power rule which says you take the exponent write it as the coefficient you leave the base exactly as it is and then reduce the exponent by one but because this is a composite function where the base of the power is a function of X we have to follow chain rule which says we then have to multiply by the derivative of the inside function so we have to multiply this by the derivative of
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the base of the power so it needs to be multiplied by G Prime of X and let me demonstrate this to you with an example in this example we're finding the derivative of a power of a function right the base of our power is a function of X it's not just X so we have to follow chain rule when using our power rule of derivatives I would start by writing the exponent as the coefficient of my power I would
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keep the base exactly as it is and then reduce this exponent by one making it to the power of three and then because the base of the power is a function of X we have a composite function meaning we now need to multiply this by the derivative of that base of the power and the derivative of 3x - 5 is 3 and then these constant factors can be multiplied together giving us 12 * 3x - 5 to the^ of 3 and I would just leave that derivative in its factored form and you'll see as we move
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throughout this video that chain rule applies to all types of composite functions not just a power of a function but it's also going to apply when we have trig functions or log functions or exponential functions so let's move on to those as well must know dtive rule number eight exponential functions for this rule if you're finding the derivative of any exponential function let's call it B to the X where B is any real number greater than zero and X is our variable which
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is in the exponent which is what makes this an exponential function when differentiating this you actually keep the exponential function exactly as it is B to the X then that has to be multiplied by the natural logarithm of the base of the power and let's also look at what happens if we make the exponent a function of X meaning we'll need to see how the chain rule applies when differentiating
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B to the power of not just X but some function of X let's call it f ofx we start by following the normal rule where we keep the exponential function exactly as it is we multiply by the natural logarithm of the base of the power but then because this is a composite function where we have a function of X in the exponent I would then have to multiply by the derivative of that exponent so this would be multiplied by frime of X and now let's do a couple examples where
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we can test out these rules all right let's have a look at these examples the first one I just want to differentiate this exponential function 3 the^ of X and I recognize it's an exponential function because our variable X is in the exponent I know when differentiating that we keep the exponential function exactly as it is 3 to the^ of X then that has to be multiplied by the natural logarithm
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of the base of the power which is three and that's it that's the derivative of 3 to the^ of X and in the second example let's look at this very special case where we're finding the derivative of e to the^ of X remember e is Oiler number it's that irrational number 2.718 and so on if I apply my derivative rule for exponential functions I know I would keep the exponential function e to the X
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and then multiply that by the natural logarithm of the base of the power so I would have the natural logarithm of e and the natural logarithm of e that means what exponent goes on E to get e well that's just one so this would be e to x * 1 which is of course just equal to e to^ of X which is a very interesting answer that the derivative of e to the^ of X is actually equal to itself the derivative of e to the x is e to ^ of X and for our final example example three let's
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differentiate 3 * e^ 2x well based on the constant multiple rule I know I can just leave that constant factor of three and then multiply that by the derivative of e to^ of 2x when differentiating e to the^ of 2x I would leave that exponential function multipli by the natural logarithm of the base and then because in the exponent we have a function of X I need to apply the chain rule which says then multiply by the derivative of that function of X the derivative of 2x
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is 2 and then I could simplify this the natural logarithm of e is 1 so I have 3 * 1 * 2 which is 6 and then I also have my exponential function e to^ of 2x so there's my simplified derivative of my original function must know derivative rule number n let's look at the rule for finding the derivative of the logarithmic function log base a of X notice for this function our variable X is in the argument of the logarithm and this a is going to be a constant value that's greater
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than zero and not one the derivative of that logarithmic function is equal to 1 / the argument X multiplied the natural logarithm of the base of the logarithm so multiplied by the natural logarithm of a and then also if we make the argument not just X but a function of X we'll need to use chain rule so if we're differentiating log base a of F ofx this would be equal to 1 / the argument which is f ofx time the natural logarithm of the base which is a but then that has to be
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multiplied by the derivative of the innermost function f ofx so this needs to be multiplied by frime of X and this frime of X can multiply into the numerator with one and now let's try a couple examples where we test out these rules let's start by looking at the first example right here where we're finding the derivative of log base 3 of X following my derivative rule I know this would equal 1 / the argument which is just x * the natural logarithm of the base of the log which is
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three and now in example two let's differentiate the natural logarithm of X following the same rule this would be equal to 1 / the argument of the logarithm X multiplied by the natural logarithm of the base of this log well this is the natural logarithm so its base is e so I'm multip L by the natural logarithm of e and the natural logarithm of e is just 1 so this answer would simplify to
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just 1 /x and in this last example we'll need to use chain rule because our logarithmic function its argument is a function of X and I also notice this constant factor of two I can just leave that out front and multiply that by the derivative of this logarithmic function the derivative of the natural log of x^2 + 1 would be 1 over the argument x^2 + 1 multiplied by the natural logarithm of the base of this log and since that's the natural logarithm its base is e and because
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the argument of the logarithm is a function of X I need to use chain rule which means multiply by the derivative of x^2 + 1 which is 2X and now simplifying my derivative I have 2 * 1 * 2x that's 4X and that's being divided by x^2 + 1 * the natural logarithm of e which is just 1 so this is being divided by x^2 + 1 and there's The Final Answer must know derivative rule number 10 the derivatives of trig functions let's look at the rules for the derivative of sinx the derivative
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of cos x and the derivative of tan X starting with the derivative of sinx the derivative of sin x is equal to cos x and the derivative of cos x is equal to sinx and the derivative of tan X is equal to secant s of X and then if any of these arguments were not just X but instead a function of X you would need to use chain rule so let's say we were doing the derivative of s not just of X but of a function of x f x you would start by knowing that the derivative of s is cosine so
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this would equal cosine of and you always leave the innermost function exactly as it is so cosine of FX but then that has to be multiplied by the derivative of the innermost function so this has to be multiplied by frime of X and now let's do a couple examples where we test out these rules starting with the first example I want the derivative of 2 cos x - 4 sin x well in differentiating this I can differentiate each of these terms separately and when doing the
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derivative of 2 cos x this constant factor of 2 gets multiplied by the derivative of cosine of x so that would be 2 * sin - 4 * the derivative of sin x which is cos x leaving me with -2 sin x - 4 cos x and then in the second example we're doing the derivative of 3 * sin of 4X this constant factor of three will just stay out front and we'll multiply that by the derivative of sin of 4X well
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the derivative of s is cosine and then I always leave the argument exactly as it is but because the argument is a function of X I have to use chain rule which tells me to multiply this whole term by the derivative of 4X and the derivative of 4X is 4 and then to simplify this I can multiply my constant factors of 3 and four together to give me 12 so my final answer is 12 cosine of 4X and before I end this video let me show you a bonus rule let me give you an example where you
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can see how implicit differentiation Works let's say you're asked to find dy/ DX for a function x^2 + y^2 = 1 notice this Y is not isolated it's not written explicitly in terms of X so instead of having to rearrange this algebraically and isolate for y you can instead do implicit differentiation and how that works you would differentiate the first term with respect to X plus I also have to differentiate this term with respect to X but it's written in terms of Y so I'll different
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it with respect to Y to start with but I really want the derivative of it with respect to X so I'll have to multiply this by dy/ DX and you can probably see why we end up with the derivative of y^2 with respect to X when we do this process and then I'll differentiate the constant one with respect to X now let me find all these derivatives the derivative of x^2 with respect to X while using the power rule that's just 2x plus the derivative of y^2 with respect to Y I can just use
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the power rule and it becomes 2 y but then don't forget there's a Dy DX beside that and that's equal to the derivative of 1 which is zero and now I just have to rearrange and isolate this equation for dy/ DX so I'll subtract the 2x to the other side and then divide the 2 Y and then simplifying this I would find that the derivative of y with respect to X is equal to X over Y and there you go
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hopefully if you made it to the end of this video you feel prepared to find find the derivative of any functions you encounter let me know in the comments what you want to top 10 of next
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