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: 9

Answer

$\dfrac{dy}{dx}=\dfrac{3x^2-6xy+2y^2}{3x^2-4yx}.$

Work Step by Step

$\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(3x^2y)+\dfrac{d}{dx}(2y^2x)=\dfrac{d}{dx}(12)\rightarrow$ $3x^2-(6xy+\dfrac{dy}{dx}(3x^2))+(\dfrac{dy}{dx}(4yx)+2y^2)=0\rightarrow$ $3x^2-6xy+2y^2=\dfrac{dy}{dx}(3x^2-4yx)\rightarrow$ $\dfrac{dy}{dx}=\dfrac{3x^2-6xy+2y^2}{3x^2-4yx}.$
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