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: 35

Answer

$y''=-\frac{4y^2+x^2}{16y^3}=-\frac{1}{4y^3}$

Work Step by Step

$x^2+4y^2=4\\ 2x+8yy'=0\\ x+4yy'=0\\ y'=\frac{-x}{4y}\\ y''=\frac{(4y)(-1)-(-x)(4y')}{(4y)^2}\\ y''=\frac{-4y+4xy'}{16y^2}\\ y''=\frac{-4y+4x(\frac{-x}{4y})}{16y^2}\\ y''=\frac{-4y-\frac{x^2}{y}}{16y^2}\\ y''=\frac{-4y^2-x^2}{16y^3}\\ y''=-\frac{4y^2+x^2}{16y^3}\\ y''=-\frac{1}{4y^3}$
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.