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 - Page 158: 30


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

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.