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: 41


$$f'(x)=\frac{\sec x(1+x\tan x)}{2\sqrt{7+x\sec x}}$$

Work Step by Step

$$f(x)=\sqrt{7+x\sec x}=(7+x\sec x)^{1/2}$$ The derivative of function $f(x)$ is $$f'(x)=\frac{1}{2}(7+x\sec x)^{-1/2}(7+x\sec x)'$$ $$f'(x)=\frac{1}{2}(7+x\sec x)^{-1/2}\Big(0+(x)'\sec x+x(\sec x)'\Big)$$ $$f'(x)=\frac{1}{2\sqrt{7+x\sec x}}(\sec x+x\sec x\tan x)$$ $$f'(x)=\frac{\sec x(1+x\tan x)}{2\sqrt{7+x\sec x}}$$
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.