Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 30

Answer

$$\frac{dy}{dx}=\frac{\cos x-3\sin x\tan(3x+1)}{\sec(3x+1)}$$

Work Step by Step

$$\frac{dy}{dx}=\frac{d}{dx}\Big(\frac{\sin x}{\sec(3x+1)}\Big)= \frac{(\sin x)'\sec(3x+1)-\sin x(\sec(3x+1))'}{\sec^2(3x+1)}= \frac{\cos x\sec(3x+1)-\sin x\sec(3x+1)\tan(3x+1)(3x+1)'}{\sec^2(3x+1)}= \frac{\cos x-\sin x\tan(3x+1)\cdot3}{\sec(3x+1)}= \frac{\cos x-3\sin x\tan(3x+1)}{\sec(3x+1)}$$
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