Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.6 Implicit Differentiation - 2.6 Exercises - Page 166: 10


For $y^5+x^2y^3=1+x^4y$ find $\frac{dy}{dx}$ $\frac{dy}{dx}=\frac{4x^3y-2xy^3}{5y^4+3x^2y^2-x^4}$

Work Step by Step

Differentiate both sides with respect to x $5y^4\frac{dy}{dx}+2xy^3+3x^2y^2\frac{dy}{dx}=4x^3y+x^4\frac{dy}{dx}$ Isolate $\frac{dy}{dx}$ $5y^4\frac{dy}{dx}+3x^2y^2\frac{dy}{dx}-x^4\frac{dy}{dx}=4x^3y-2xy^3$ $\frac{dy}{dx}(5y^4+3x^2y^2-x^4)=4x^3y-2xy^3$ $\frac{dy}{dx}=\frac{4x^3y-2xy^3}{5y^4+3x^2y^2-x^4}$
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.