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

1 Answer

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}