Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 30

Answer

$J'(\theta)=2n\sec^{2}(n\theta)\tan(n\theta)$

Work Step by Step

$J(\theta)=\tan^{2}(n\theta)$ (Here, $n$ is a constant) Differentiate using the chain rule: $J'(\theta)=2\tan(n\theta)[\tan(n\theta)]'=...$ Apply the chain rule one more time to find $[\tan(n\theta)]'$ $...=2\tan(n\theta)[\sec^{2}(n\theta)](n\theta)'=2n\sec^{2}(n\theta)\tan(n\theta)$
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