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 - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 425: 24


$y' = \dfrac{2\sin(\ln x) \cos(\ln x)}{x}$

Work Step by Step

In order to derivate this function you have to apply the chain rule Let's make a «u» substitution to make it easier $f(u) = u^2$ $u = \sin(\ln x)$ Derivate the function: $f'(u) = 2uu'$ Now let's find u' *Note: Here you have to apply the chain rule again $u' = \dfrac{\cos(\ln x)}{x}$ Then undo the substitution, simplify and get the answer: $y' = \dfrac{2\sin(\ln x) \cos(\ln x)}{x}$
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