How do you use the first and second derivatives to sketch # y=(x+2)/(x-1)#?

1 Answer
Harish Chandra Rajpoot
Jun 24, 2018

#y'=-3/(x-1)^2#, #y''=6\(x-1)^3#

Explanation:

Given: #y=\frac{x+2}{x-1}#

#\frac{dy}{dx}=\frac{(x-1)\frac{d}{dx}(x+2)-(x+2)\frac{d}{dx}(x-1)}{(x-1)^2}#
#\frac{dy}{dx}=\frac{(x-1)\cdot1-(x+2)\cdot1}{(x-1)^2}#
#\frac{dy}{dx}=\frac{-3}{(x-1)^2}#
#\frac{d^2y}{dx^2}=-3\frac{d}{dx}\frac{1}{(x-1)^2}#
#\frac{d^2y}{dx^2}=-3\cdot\frac{-2}{(x-1)^3}#
#\frac{d^2y}{dx^2}=\frac{6}{(x-1)^3}#

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