Calculus 8th Edition

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

Chapter 2 - Derivatives - Review - Exercises - Page 197: 40


$$y'=\frac{-\pi\sin (\cos \sqrt{\sin \pi x} ) [\sin \sqrt{\sin \pi x}]\cos \pi x}{ \sqrt{\sin \pi x}}$$

Work Step by Step

Given $$ y = \sin^2 (\cos \sqrt{\sin \pi x} ) $$ Then \begin{align*} y'& =2\sin (\cos \sqrt{\sin \pi x} ) [-\sin \sqrt{\sin \pi x}]\frac{1}{2 \sqrt{\sin \pi x}}(\pi\cos \pi x)\\ &=\frac{-\pi\sin (\cos \sqrt{\sin \pi x} ) [\sin \sqrt{\sin \pi x}]\cos \pi x}{ \sqrt{\sin \pi x}} \end{align*}
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