Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 432: 26


$y' = -\dfrac{e^x\sin (e^x)}{\cos (e^x)}$

Work Step by Step

In order to derivate this function you have to apply the chain rule Let's make an «u» substitution to make it easier $f(u) = \ln(u)$ $u = \cos(e^x)$ Derivate the function: $f'(u) = \dfrac{u'}{u}$ Now let's find u' $u'= e^x\sin (e^x)$ Then undo the substitution, simplify and get the answer: $f'(x) = -\dfrac{e^x\sin (e^x)}{\cos (e^x)}$
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