Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6 - Page 158: 53



Work Step by Step

$$\frac{dy}{dx}=\big(\frac{1+x}{1-x}\Big)'= \frac{(1+x)'(1-x)-(1+x)(1-x)'}{(1-x)^2}= \frac{1-x-(1+x)\cdot(-1)}{(1-x)^2}=\frac{2}{(1-x)^2}$$ $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{dy}{dx}\Big)=\Big(\frac{2}{(1-x)^2}\Big)'=(2(1-x)^{-2})'=2\cdot(-2)(1-x)^{-3}(1-x)'=-\frac{4}{(1-x)^3}\cdot(-1)=\frac{4}{(1-x)^3}$$
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.