Simplify the Difference Quotient when $f(x)= \frac{15}{x^2}$ Simplify the Difference Quotient when $f(x)= \frac{15}{x^2}$. The difference quotient formula is $\frac{f(x+h)- f(x)}{ h}$.
I understand how to simplify basic difference quotient equations. In this case, you start with the formula: $\displaystyle\frac{\frac{15}{{(x+h)}^2}-\frac{15}{x^2}}{h}$. The answer to the problem is $\displaystyle\frac{-15(h+2x)}{x^2(h+x)^2}$.
I have tried multiplying by the conjugate and the Least Common Denominator, but my algebra skills aren't leading me to the correct answer. Please help!

$f(x)=\frac{15}{x^2}$
Now 
$\frac{f(x+h)-f(x)}{h}=\frac{\frac{15}{(x+h)^2}-\frac{15}{x^2}}{h}$
$=\frac{15}{h}[\frac{1}{(x+h)^2}-\frac{1}{x^2}]$
$=\frac{15}{h}[\frac{x^2-(x+h)^2}{x^2(x+h)^2}]$
$=\frac{15}{h}[\frac{x^2-x^2-2xh-h^2}{x^2(x+h)^2}]$
$=\frac{-15}{h}[\frac{2xh+h^2}{x^2(x+h)^2}]$
$=-15[\frac{2x+h}{x^2(x+h)^2}]$
