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: 4

Answer

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

Work Step by Step

$\dfrac{d}{dx}(2x^3)+\dfrac{d}{dx}(3y^3)=\dfrac{d}{dx}(64)\rightarrow$ $6x^2+\dfrac{dy}{dx}(9y^2)=0\rightarrow$ $\dfrac{dy}{dx}=-\dfrac{6x^2}{9y^2}=-\dfrac{2x^2}{3y^2}.$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.