Answer
$-\displaystyle \frac{2x+h}{x^{2}(x+h)^{2}}$
Work Step by Step
First, simplify $\displaystyle \frac{1}{x+h}-\frac{1}{x}$
LCD = $x^{2}(x+h)^{2}$
$\displaystyle \frac{1}{(x+h)^{2}}\cdot\frac{x^{2}}{x^{2}}-\frac{1}{x^{2}}\cdot\frac{(x+h)^{2}}{(x+h)^{2}}=\frac{x^{2}-(x+h)^{2}}{x^{2}(x+h)^{2}}$
$=\displaystyle \frac{x^{2}-(x^{2}+2hx+h^{2})}{x^{2}(x+h)^{2}}$
$=\displaystyle \frac{-2hx-h^{2}}{x^{2}(x+h)^{2}}$
$=\displaystyle \frac{-h(2x+h)}{x^{2}(x+h)^{2}}$
Now,
$\displaystyle \frac{1}{h}\left[\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}\right]$ = $\displaystyle \frac{1}{h}\left[\frac{-h(2x+h)}{x^{2}(x+h)^{2}}\right]$
$... $ h cancels
$=\displaystyle \frac{-(2x+h)}{x^{2}(x+h)^{2}}$
$=-\displaystyle \frac{2x+h}{x^{2}(x+h)^{2}}$
