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: 25

Answer

$$h'(\theta)=\frac{\theta(2\sin\theta+\theta\cos\theta)}{\cos^2(\theta^2\sin\theta)}$$

Work Step by Step

$$h'(\theta)=(\tan(\theta^2\sin\theta))'=\frac{1}{\cos^2(\theta^2\sin\theta)}(\theta^2\sin\theta)'=\frac{1}{\cos^2(\theta^2\sin\theta)}((\theta^2)'\sin\theta+\theta^2(\sin\theta)')= \frac{1}{\cos^2(\theta^2\sin\theta)}(2\theta\sin\theta+\theta^2\cos\theta)=\frac{\theta(2\sin\theta+\theta\cos\theta)}{\cos^2(\theta^2\sin\theta)}$$
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