Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 215: 16



Work Step by Step

$x^2y^2=x^2+y^2\\ \frac{dy}{dx}(x^2y^2=x^2+y^2)\\ 2xy^2+2yx^2\frac{dy}{dx}=2x+2y\frac{dy}{dx}\\ 2xy^2-2x=2y\frac{dy}{dx}-2yx^2\frac{dy}{dx}\\ 2xy^2-2x=\frac{dy}{dx}(2y-2yx^2)\\ \frac{dy}{dx}=\frac{2xy^2-2x}{2y-2yx^2}\\ \frac{dy}{dx}=\frac{2(xy^2-x)}{2(y-yx^2)}\\ \frac{dy}{dx}=\frac{xy^2-x}{y-yx^2}$
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