Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.6 Implicit Differentiation - 2.6 Exercises - Page 166: 8


For $x^3-xy^2+y^3=1$ find $\frac{dy}{dx}$

Work Step by Step

Differentiate both side with respect to x $3x^2-y^2-2xy\frac{dy}{dx}+3y^2\frac{dy}{dx} =0$ Isolate $\frac{dy}{dx}$ $\frac{dy}{dx}(3y^2-2xy)=y^2-3x^2$ $\frac{dy}{dx}=\frac{y^2-3x^2}{(3y^2-2xy)}$
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