Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 57


$$ y' =- \frac{1}{(z-1)^2}\sqrt{\frac{z-1}{z+1}}.$$

Work Step by Step

Recall that $(x^n)'=nx^{n-1}$ Since $ y=\sqrt{\frac{z+1}{z-1}}$, by using the chain and product rules, the derivative $ y'$ is given by $$ y'=\frac{1}{2\sqrt{\frac{z+1}{z-1}}}\left( \frac{(z-1)(1)-(z+1)(1)}{(z-1)^2}\right)\\ =- \frac{1}{(z-1)^2}\sqrt{\frac{z-1}{z+1}}.$$
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.