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: 38

Answer

$$\frac{dy}{dx}=\frac{68x(1+x^2)^{16}}{(1-x^2)^{18}}$$

Work Step by Step

$$\frac{dy}{dx}=\Big(\Big(\frac{1+x^2}{1-x^2}\Big)^{17}\Big)'=17\Big(\frac{1+x^2}{1-x^2}\Big)^{16}\Big(\frac{1+x^2}{1-x^2}\Big)'= 17\Big(\frac{1+x^2}{1-x^2}\Big)^{16}\frac{(1+x^2)'(1-x^2)-(1+x^2)(1-x^2)'}{(1-x^2)^2}= 17\Big(\frac{1+x^2}{1-x^2}\Big)^{16}\frac{2x(1-x^2)-(1+x^2)\cdot(-2x)}{(1-x^2)^2}= \frac{68x(1+x^2)^{16}}{(1-x^2)^{18}}$$
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