Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 178: 47



Work Step by Step

Take the derivative of the equation on each side separately. Apply chain rule when differentiating the "y" variables since we are differentiating with respect to x: $2y\frac{dy}{dx}=\frac{(x+1)(1)-(x)(1)}{(x+1)^2}$ $2y\frac{dy}{dx}=\frac{x+1-x}{(x+1)^2}$ $2y\frac{dy}{dx}=\frac{1}{(x+1)^2}$ Move all terms with dy/dx to one side of the equation, and isolate dy/dx: $\frac{dy}{dx}=\frac{1}{2y(x+1)^2}$
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