University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.7 - Implicit Differentiation - Exercises - Page 164: 7



Work Step by Step

$$y^2=\frac{x-1}{x+1}$$ 1) Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(y^2)=\frac{d}{dx}\Big(\frac{x-1}{x+1}\Big)$$ $$2y\frac{dy}{dx}=\frac{(x+1)-(x-1)}{(x+1)^2}$$ $$2y\frac{dy}{dx}=\frac{x+1-x+1}{(x+1)^2}$$ $$2y\frac{dy}{dx}=\frac{2}{(x+1)^2}$$ $$y\frac{dy}{dx}=\frac{1}{(x+1)^2}$$ 2) Now solve for $dy/dx$: $$\frac{dy}{dx}=\frac{1}{y(x+1)^2}$$
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