Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 32

Answer

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

Work Step by Step

$$J'(\theta)=(\tan^2(n\theta))'=2\tan(n\theta)(\tan(n\theta))'=2\tan(n\theta)\frac{1}{\cos^2(n\theta)}(n\theta)'= 2\tan(n\theta)\frac{1}{\cos^2(n\theta)}\cdot n=2n\tan(n\theta)\sec^2(n\theta)$$
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