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: 62

Answer

$$\frac{dy}{dt}=-\frac{5}{3}\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\cos\Big(\frac{t}{3}\Big)$$

Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}\Big(\cos\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\Big)$$ Following the Chain Rule: $$\frac{dy}{dt}=-\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\frac{d}{dt}\Big(5\sin\Big(\frac{t}{3}\Big)$$ $$\frac{dy}{dt}=-\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\times5\cos\Big(\frac{t}{3}\Big)\frac{d}{dt}\Big(\frac{t}{3}\Big)$$ $$\frac{dy}{dt}=-5\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\cos\Big(\frac{t}{3}\Big)\Big(\frac{1}{3}\Big)$$ $$\frac{dy}{dt}=-\frac{5}{3}\sin\Big(5\sin\Big(\frac{t}{3}\Big)\Big)\cos\Big(\frac{t}{3}\Big)$$
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.