Answer
$ \displaystyle \frac{2}{x^{3}}+\frac{1}{x^{2}}$
Work Step by Step
$f''(x)=\displaystyle \frac{d}{dx}\left[\frac{df}{dx}\right]=\frac{d^{2}f}{dx^{2}}$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[x^{-1}-\ln x]=-x^{-2}-\frac{1}{x}$
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}[ -x^{-2}-x^{-1}]=-(-2x^{-3})-(-x^{-2})$
= $ \displaystyle \frac{2}{x^{3}}+\frac{1}{x^{2}}$
