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Numerade Educator

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Problem 63 Hard Difficulty

Find the derivative of the function.

$ y = \displaystyle \int^{\sin x}_{\cos x} \ln (1 + 2v) \, dv $

Answer

$\sin x \ln (1+2 \cos x)+\cos x \ln (1+2 \sin x)$

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SK

Sam K.

May 25, 2020

do you not need to find the antiderivative of ln(1+2v) ?

Video Transcript

for this problem where to find the derivative of Y equal to the integral firm cosine X two syntax of the natural log of one plus T V D V. Not to do this. You have to apply fundamental theorem of calculus which states that the derivative of the integral from A to let's say X of F F T D T. This is equal to F of X. Now in our given function the bands are both functions of X. So we need to partition this interval into two intervals in which one of the endpoint is a constant. That's e knows that in the interval co sign X to cynics there is some value A in between it. So we can say that this is equal to um the interval from co sign X to A. And then union from A to sign of X. And so we can rewrite the function into why that's equal to the integral from cosign X to E. Of L N. Of one plus two V. D V plus we have the integral from a two Synnex of L N. Of one plus T V D. V. And because you're following fundamental theorem of calculus, we need to we write the first integral so that the function the boundary function will go up and then the boundary they will be the bottom boundary. And so we rewrite this into why is just equal to the negative of the integral from a to co sign X. Of the natural log of one plus two Vdv plus we have the integral from A to cynics, ah the natural log of one plus two V. D. V. And so by fundamental theorem of calculus and chain rule we have why prime? This is equal to the negative of the natural log of one plus two times cosine x times the derivative of cosine X. And then plus we have um Ln of one plus two times cynics times the derivative of Sine X. And simplifying this, we get negative Ln of one plus to cosign X times the derivative Goldstein exit just negative Synnex plus we have Ln of one plus two syntax times the derivative of Sine X. Which is co sign X. And so we get sin X times Ln of one plus to co sign X plus. Co sign X times Ln of one plus to sign Mhm. And so this is the derivative of the function.