Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises - Page 158: 39


$$f'(t)=-\sec^2(\sec(\cos t))\sec(\cos t)\tan(\cos t)\sin t$$

Work Step by Step

$$f'(t)=(\tan(\sec(\cos t)))'=\frac{1}{\cos^2(\sec(\cos t))}(\sec(\cos t))'= \sec^2(\sec(\cos t))\left(\frac{1}{\cos(\cos t)}\right)'= \sec^2(\sec(\cos t))(-\frac{1}{\cos^2(\cos t)})(\cos(\cos t))'= -\sec^2(\sec(\cos t))\frac{1}{\cos^2(\cos t))}(-\sin(\cos t))(\cos t)'= \sec^2(\sec(\cos t))\sec(\cos t)\tan(\cos t)(-\sin t)= -\sec^2(\sec(\cos t))\sec(\cos t)\tan(\cos t)\sin t$$
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.