Calculus 8th Edition

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

Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 23


$$2\sec^22\theta e^{\tan 2\theta }$$

Work Step by Step

Given $$h(\theta )=e^{\tan 2\theta }$$ Then \begin{align*} h'(\theta)&=\frac{d}{d\theta}\left(e^{\tan 2\theta }\right) \\ &=e^{\tan 2\theta }\frac{d}{d\theta}\left( {\tan 2\theta }\right) \\ &=2\sec^22\theta e^{\tan 2\theta } \end{align*}
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