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 167: 36



Work Step by Step

$x^2+xy+y^2=3\\ 2x+xy'+y+2yy'=0\\ xy'+2yy'=-2x-y\\ y'=\frac{-2x-y}{x+2y}\\ y''=\frac{(x+2y)(-2-y')-(-2x-y)(1+2y')}{(x+2y)^2}\\ y''=\frac{-2x-4y-xy'-2yy'+2x+4xy'+y+2yy'}{(x+2y)^2}\\ y''=\frac{3xy'-3y}{(x+2y)^2}\\ y''=\frac{3x(\frac{-2x-y}{x+2y})-3y}{(x+2y)^2}\\ y''=\frac{\frac{-6x^2-3xy}{x+2y}-3y}{(x+2y)^2}\\ y''=\frac{\frac{-6x^2-3xy-3xy-6y^2}{x+2y}}{(x+2y)^2}\\ y''=-\frac{6(x^2+xy+y^2)}{(x+2y)^3}\\ y''=-\frac{18}{(x+2y)^3}$
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.