Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 197: 10

Answer

$g'\left( \theta \right) = \left( {{{\tan }^2}\theta + \tan \theta - \theta } \right){e^\theta }$

Work Step by Step

$$\eqalign{ & g\left( \theta \right) = {e^\theta }\left( {\tan \theta - \theta } \right) \cr & {\text{Differentiating}} \cr & g'\left( \theta \right) = \frac{d}{{d\theta }}\left[ {{e^\theta }\left( {\tan \theta - \theta } \right)} \right] \cr & {\text{Use the product rule for derivatives }}\left( {fh} \right)' = fh' + hf' \cr & {\text{Let }}f = {e^\theta }{\text{ and }}h = \tan \theta - \theta ,{\text{ then}} \cr & g'\left( \theta \right) = {e^\theta }\frac{d}{{d\theta }}\left[ {\tan \theta - \theta } \right] + \left( {\tan \theta - \theta } \right)\frac{d}{{d\theta }}\left[ {{e^\theta }} \right] \cr & \cr & {\text{Use the Derivatives formulas }} \cr & \frac{d}{{dx}}\left[ {\tan x} \right] = {\sec ^2}x,{\text{ }}\frac{d}{{dx}}\left[ {{e^x}} \right] = {e^x} \cr & {\text{therefore}} \cr & g'\left( \theta \right) = {e^\theta }\left( {{{\sec }^2}\theta - 1} \right) + \left( {\tan \theta - \theta } \right){e^\theta } \cr & {\text{Factoring}} \cr & g'\left( \theta \right) = \left( {{{\sec }^2}\theta - 1 + \tan \theta - \theta } \right){e^\theta } \cr & g'\left( \theta \right) =(\tan^2\theta+\tan\theta-\theta)e^{\theta}} $$
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