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: 52


$$\frac{dy}{dt}==2\pi\sec^2\pi t\tan\pi t$$

Work Step by Step

According to the Chain Rule: $$\frac{dy}{dt}=\frac{d}{dt}(\sec^2\pi t)$$ $$\frac{dy}{dt}=2\sec\pi t\frac{d}{dt}(\sec \pi t)$$ $$\frac{dy}{dt}=2\sec\pi t(\sec\pi t\tan\pi t)\frac{d}{dt}(\pi t)$$ $$\frac{dy}{dt}=2\sec^2\pi t\tan\pi t\times(\pi)=2\pi\sec^2\pi t\tan\pi t$$
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.