Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 39


$$y' =\cot x -\sin x\cos x$$

Work Step by Step

Given $$y=\ln\left(\sin x\right)-\frac{1}{2}\sin^2x$$ Then \begin{align*} y'&=\frac{d}{dx}\left(\ln \left(\sin \left(x\right)\right)\right)-\frac{d}{dx}\left(\frac{1}{2}\sin ^2\left(x\right)\right)\\ &=\frac{\cos x}{\sin x} -\sin x\cos x\\ &=\cot x -\sin x\cos x \end{align*}
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