University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 203: 91



Work Step by Step

$y^3+y=2cosx$ Differentiate the above equation with respect to x: $(3y^2+1)\frac{dy}{dx}=-2\sin{x}$ on taking double differentiation $(3y^2+1)\frac{d^2y}{dx}+(6y)\frac{dy}{dx}=-2\cos{x}$ at point$(0,1)$ $(3+1)\frac{dy}{dx}=-2\sin{0}=0$ $\frac{dy}{dx}=0$ and $(3+1)\frac{d^2y}{dx}+(6)\times0=-2\cos{0}$ $\frac{d^2y}{dx}=\frac{-1}{2}$
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