University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 174: 16

Answer

$$y'=\cot x$$

Work Step by Step

$$y=\ln(\sin x)$$ Recall the Derivative Rule for Natural Logarithm: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$ Therefore, we have $$y'=\Big(\ln\sin x\Big)'=\frac{1}{\sin x}(\sin x)'=\frac{\cos x}{\sin x}$$ $$y'=\cot x$$
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