University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 158: 42



Work Step by Step

$$g(x)=\frac{\tan 3x}{(x+7)^4}$$ The derivative of $g(x)$ is $$g'(x)=\frac{(\tan3x)'(x+7)^4-\tan3x\Big((x+7)^4\Big)'}{(x+7)^8}$$ According to the Chain Rule, we have $(\tan3x)'=\sec^23x(3x)'=3\sec^23x$ and $\Big((x+7)^4\Big)'=4(x+7)^3(x+7)'=4(x+7)^3$ Therefore, $$g'(x)=\frac{3\sec^23x(x+7)^4-4\tan3x(x+7)^3}{(x+7)^8}$$ $$g'(x)=\frac{(x+7)^3\Big(3\sec^23x(x+7)-4\tan3x\Big)}{(x+7)^8}$$ $$g'(x)=\frac{3\sec^23x(x+7)-4\tan3x}{(x+7)^5}$$
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.