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: 68



Work Step by Step

$$\frac{dy}{dt}=\frac{d}{dt}\cos^4(\sec^23t)$$ Following the Chain Rule: $$\frac{dy}{dt}=4\cos^3(\sec^23t)\frac{d}{dt}(\cos(\sec^23t))$$ $$\frac{dy}{dt}=4\cos^3(\sec^23t)(-\sin(\sec^23t))\frac{d}{dt}(\sec^23t)$$ $$\frac{dy}{dt}=4\cos^3(\sec^23t)(-\sin(\sec^23t))2\sec3t\frac{d}{dt}(\sec3t)$$ $$\frac{dy}{dt}=8\cos^3(\sec^23t)(-\sin(\sec^23t))\sec3t(\sec3t\tan3t\frac{d}{dt}(3t))$$ $$\frac{dy}{dt}=-24\cos^3(\sec^23t)\sin(\sec^23t)\sec^23t\tan3t$$
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.