Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises: 6

Answer

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

Work Step by Step

$g(\theta)=e^{\theta}(\tan\theta-\theta)$ Differentiate using the product rule: $g'(\theta)=(e^{\theta})(\tan\theta-\theta)'+(\tan\theta-\theta)(e^{\theta})'=...$ $...=(e^{\theta})(\sec^{2}\theta-1)+(\tan\theta-\theta)(e^{\theta})=...$ We can use the following identity: $\sec^{2}\theta-1=\tan^{2}\theta$: $...=e^{\theta}\tan^{2}\theta+(\tan\theta-\theta)e^{\theta}=...$ Take out common factor $e^{\theta}$ for a better looking answer: $...=e^{\theta}(\tan^{2}\theta+\tan\theta-\theta)$
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