Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 7


$$\frac{d}{dx}\frac{y}{y+1}= \frac{1}{(y+1)^2}\frac{dy}{dx}.$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Using the quotient and chain rules, we have $$\frac{d}{dx}\frac{y}{y+1}=\frac{(y+1)\frac{dy}{dx}-y\frac{dy}{dx}}{(y+1)^2}= \frac{1}{(y+1)^2}\frac{dy}{dx}.$$
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