Answer
$\dfrac{dy}{dx}=-\dfrac{2x^2}{3y^2}.$
Work Step by Step
$\dfrac{d}{dx}(2x^3)+\dfrac{d}{dx}(3y^3)=\dfrac{d}{dx}(64)\rightarrow$
$6x^2+\dfrac{dy}{dx}(9y^2)=0\rightarrow$
$\dfrac{dy}{dx}=-\dfrac{6x^2}{9y^2}=-\dfrac{2x^2}{3y^2}.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 2 - Differentiation - 2.5 Exercises - Page 145: 4
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