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: 45



Work Step by Step

$$r=\sin(\theta^2)\cos(2\theta)$$ The derivative of $r$ is $$r'=\Big(\sin(\theta^2)\Big)'\cos(2\theta)+\sin(\theta^2)\Big(\cos(2\theta)\Big)'$$ According to the Chain Rule, we have $\Big(\sin(\theta^2)\Big)'=\cos(\theta^2)(\theta^2)'=2\theta\cos(\theta^2)$ $\Big(\cos(2\theta)\Big)'=-\sin(2\theta)(2\theta)'=-2\sin(2\theta)$ Therefore, $$r'=2\theta\cos(\theta^2)\cos(2\theta)+\sin(\theta^2)\Big(-2\sin(2\theta)\Big)$$ $$r'=2\theta\cos(\theta^2)\cos(2\theta)-2\sin(\theta^2)\sin(2\theta)$$
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.