derivatives-test-solutions-part-i-3

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Right students is go ahead and take a look at Tom the remaining review problems on the chain rule so for question number four we are to find F prime of X is F of X is equal to sign square two X minus five divided by X square times the natural log of them of X so we can notice here that we have quotient of two functions let's call the numerator function you and the denominator function the to how to find the derivative of the quotient of two functions so that the have you over be right what will the derivative the so is simply going to use the quotient rule okay so the quotient rule it's we know that is the you prime minus you UV prime divided by square okay so that's the quotient rule right so all you have to do here is fine you prime in v prime and then substitute it into this formula and that will give us the result that we desire okay let's start with you to you is equal to sign square of two X minus five now this is a composite function is a composition of three functions okay we have the square function the sign derivative function and in the linear function X minus five so we can rights you as F of G of H of X okay F of G of each of X so this is F of G of H of X and you prime applying the chain rule to the F prime of G of H of X times G prime of H of X times each prime of X okay so let's go ahead and of the compose you Indian we can find the derivative of the different functions in plug it into the general formula okay so use F of G of H of X which is equal to all sine square of two X minus five right taking how the F so F of X is X square then G of X actually G of HMX is going to be sine of two X minus five we can see how this function can't result a result of the composition of these two okay with ethology G of X is sine X and on the limits is of H of X the innermost function is all two X minus five okay now let's go ahead and find the derivative of these three functions F prime of X is to X find the power rule you prime of X using trigonometric differentiation is cosine X and then H prime of X is simply to okay now what we going to do is obliges resulting Got to all the chain rule formula right so U prime is going to be at prime of G of H of X is together that simply take G roots G I'm sorry G of H of X limit take that okay here okay and then to get this to that in the U prime is going to be all to the X is getting two times sine of two X minus five right to that's F prime of G of H of X times G prime of H of X is simply take H and plug it into G prime of X so that yields is the cosine X will have cosine two X minus five times H prime of X which is to our right side is the oldest cases into this chain rule formula to generate this expression right here okay have one piece of the chain rule formula the U prime now let's go ahead and of find the product okay so we remember what the is the is X square times the natural logarithm of X okay so let's write that down the have of the equals X square by the natural log of them of X so we have the product of two functions here so we going to be using the product rule okay so let's call this expression going to be U and then allwill be the remember what the product rule is UV prime is equal to the U prime plus U the prime let's apply that here we going to have the the natural logarithm of X times the derivative of X square U prime is going to be two X plus U which is X square times the derivative of the the prime the is on the natural of them of X so be prime is one over X right so let's simplify that is simply going to be two X times the natural logarithm of X plus X okay so U reduce these two so will simply put everything together F prime of X is simply going to be the which is X square of the natural of them of X times U prime U prime is this expression right here all we can multiply to into just the fourth for sine two X minus five cosine two X minus five this i directories s U the prime minus I'm sorry this is the U prime minus UV prime okay so U is sine square times two X minus five the prime is this result right here two X times the natural of them of X let's X and in this entire expression according to the quotient rule is the be divided by the square so be is that so we just have X where LN X raised to the second power okay so this is the desired results on for F prime of X prime is going to take a look at another question right so we are to find the why the X of this problem now we going to have to using chain rule here this is a composition of two functions so why is equal to F of G of X okay why prime do I the X is equal to F prime the derivative of the outside function about we can at the inner function times the derivative of the inner function okay now let's go ahead and decomposed our functions so we know what

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