Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 216: 38



Work Step by Step

Find y' by taking derivative of both sides of the equation: $3x^2-3y^2\times dy/dx=0$ $y'=\frac{x^2}{y^2}$ Find y": $y"=\frac{y^2(2x)-(x^2)(2y\times y')}{y^4}$ Plug y' into y": $y"=\frac{2xy^2-2yx^2\times \frac{x^2}{y^2}}{y^4}$ Simplify: $y"=\frac{2xy^4-2yx^4}{y^6}=\frac{2xy^3-2x^4}{y^5}$ Substitute $y^3=x^3-7$ and simplify: $y"=\frac{-14x}{y^5}$
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.