Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 48

Answer

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

Work Step by Step

$xy-1=2x+y^2$ $\frac{d}{dx}[xy-1]=\frac{d}{dx}[2x+y^2]$ $y+x\frac{dy}{dx}=2+2y\frac{dy}{dx}$ $(x-2y)\frac{dy}{dx}=2-y$ $\frac{dy}{dx}=\frac{2-y}{x-2y}$ $\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{dy}{dx}]$ $\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{2-y}{x-2y}]$ $\frac{d^2y}{dx^2}=\frac{-\frac{dy}{dx}(x-2y)-(2-y)(1-2\frac{dy}{dx})}{(x-2y)^2}$ $\frac{d^2y}{dx^2}=\frac{-(\frac{2-y}{x-2y})(x-2y)-(2-y)(1-2(\frac{2-y}{x-2y}))}{(x-2y)^2}$ $\frac{d^2y}{dx^2}=\frac{-(2-y)-(2-y)(1-2(\frac{2-y}{x-2y}))}{(x-2y)^2}$ $\frac{d^2y}{dx^2}=\frac{-(2-y)-(2-y)(1-\frac{4-2y}{x-2y})}{(x-2y)^2}$ $\frac{d^2y}{dx^2}=\frac{y-2+(y-2)(1-\frac{4-2y}{x-2y})}{(x-2y)^2}$ $\frac{d^2y}{dx^2}=\frac{(y-2)(1+1-\frac{4-2y}{x-2y})}{(x-2y)^2}$ $\frac{d^2y}{dx^2}=\frac{(y-2)(2-2\frac{2-y}{x-2y})}{(x-2y)^2}$ $\frac{d^2y}{dx^2}=\frac{2(y-2)(1-\frac{2-y}{x-2y})}{(x-2y)^2}$ $\frac{d^2y}{dx^2}=\frac{2(y-2)((x-2y)-(2-y))}{(x-2y)^3}$ $\frac{d^2y}{dx^2}=\frac{2(y-2)((x-2y)+y-2)}{(x-2y)^3}$ $\frac{d^2y}{dx^2}=\frac{2(y-2)(x-y-2)}{(x-2y)^3}$
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