Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.8 Implicit Differentiation - 3.8 Exercises - Page 200: 32


$\frac{d^2y}{dx^2} = \frac{-2y^2-4x^2}{y^3}$

Work Step by Step

$2x^2+y^2=4$ First Derivative: $4x+2y\frac{dy}{dx}=0$ $2y\frac{dy}{dx} = -4x$ $\frac{dy}{dx} = \frac{-2x}{y}$ Second Derivative: $\frac{d^2y}{dx^2} = \frac{(-2)(y)-(-2x)(\frac{dy}{dx})}{y^2} = \frac{-2y-\frac{4x^2}{y}}{y^2} = \frac{-2y^2-4x^2}{y^3}$
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.