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: 9

Answer

For $\frac{x^2}{x+y}=y^2+1$ find $\frac{dy}{dx}$ $\frac{dy}{dx}=\frac{x^2+2xy}{(x+y)^22y+x^2}$

Work Step by Step

Differentiate both sides with respect to x $\frac{2x(x+y)-x^2(1+\frac{dy}{dx})}{(x+y)^2}=2y\frac{dy}{dx}$ Isolate $\frac{dy}{dx}$ $2x^2+2xy-x^2-x^2\frac{dy}{dx}=\frac{dy}{dx}(x+y)^22y$ $x^2+2xy=\frac{dy}{dx}(x+y)^22y+x^2\frac{dy}{dx}$ $x^2+2xy=\frac{dy}{dx}((x+y)^22y+x^2)$ $\frac{dy}{dx}=\frac{x^2+2xy}{(x+y)^22y+x^2}$
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