Calculus 8th Edition

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

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 47

Answer

$$y'=\frac{-3 \sin \left(e^{\sqrt{\tan 3 x}}\right) e^{\sqrt{\tan 3 x}} \sec ^{2}(3 x)}{2 \sqrt{\tan 3 x}} $$

Work Step by Step

Given $$y=\cos(e^{\sqrt{\tan 3x}})$$ Then \begin{align*} y^{\prime} &=-\sin \left(e^{\sqrt{\tan 3 x}}\right) \cdot\left(e^{\sqrt{\tan 3 x}}\right)^{\prime}\\ &=-\sin \left(e^{\sqrt{\tan 3 x}}\right) e^{\sqrt{\tan 3 x}} \cdot \frac{1}{2}(\tan 3 x)^{-1 / 2} \cdot \sec ^{2}(3 x) \cdot 3 \\ &=\frac{-3 \sin \left(e^{\sqrt{\tan 3 x}}\right) e^{\sqrt{\tan 3 x}} \sec ^{2}(3 x)}{2 \sqrt{\tan 3 x}} \end{align*}
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