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: 31


$\frac{d^2x}{dy^2} = -\frac{-1}{4y^3}$

Work Step by Step

$x+y^2=1$ Take Derivative and Solve for $\frac{dy}{dx}$ $1+2y\frac{dy}{dx}=0$ $\frac{dy}{dx} = \frac{-1}{2y}$ Take Second Derivative and Solve for $\frac{d^2y}{dx^2}$ $\frac{d^2x}{dy^2} = \frac{2}{4y^2}\times \frac{-1}{2y} = -\frac{-1}{4y^3}$
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