Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 6

Answer

$\dfrac{dy}{dx}=\dfrac{2xy+y^2}{-x^2-2yx}.$

Work Step by Step

$\dfrac{d}{dx}(x^2y)+\dfrac{d}{dx}(y^2x)=\dfrac{d}{dx}(-2)\rightarrow$ $(2xy+\dfrac{dy}{dx}(x^2))+(y^2+\dfrac{dy}{dx}(2yx))=0\rightarrow$ $2xy+y^2=\dfrac{dy}{dx}(-x^2-2yx)\rightarrow$ $\dfrac{dy}{dx}=\dfrac{2xy+y^2}{-x^2-2yx}.$
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