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: 7

Answer

$\dfrac{dy}{dx}=\dfrac{3y^3x^2-1}{1-3x^3y^2}.$

Work Step by Step

$\dfrac{d}{dx}(x^3y^3)-\dfrac{d}{dx}(y)=\dfrac{d}{dx}(x)\rightarrow$ $(3x^2y^3+\dfrac{dy}{dx}(3y^2x^3))-\dfrac{dy}{dx}=1\rightarrow$ $3y^3x^2-1=\dfrac{dy}{dx}(1-3x^3y^2)\rightarrow$ $\dfrac{dy}{dx}=\dfrac{3y^3x^2-1}{1-3x^3y^2}.$
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