Finding the derivative of $f(x) = \frac{8}{\sqrt{x -2}}$ using first principles. How would you go about determining the derivative of ( $f(x) = \frac{8}{\sqrt{x -2}}$ ) using the limit definition of the derivative 
(i.e. $\lim\limits_{h\to 0} = \frac{f(x+h) - f(x)}{h}$) as opposed to just applying the chain rule. So I'm thinking this counts as an algebra question but can't find too many examples dealing with simplification of polynomial expressions with fractional exponents. 
I got as far as, 
$\frac {1}{h}  (\frac{8}{\sqrt{x+h-2}} - \frac{8}{\sqrt{x-2}})$
In case I'm not using correct terminology or being unclear, what I mean is, can one algebraically eliminate the $h$ from the denominator in the above expression to take the value of the limit? 

Hint: For $x>2$ and any non-zero $h$ such that $x+h>2$:
$$\frac {1}{h}  \left(\frac{8}{\sqrt{x+h-2}} - \frac{8}{\sqrt{x-2}}\right)$$
$$ =  \frac {1}{h}  \frac{8(\sqrt{x-2} - \sqrt{x+h-2})}{\sqrt{x+h-2}\sqrt{x-2}} $$
$$ =   \frac {1}{h}  \frac{8[(x-2) - (x+h-2)]}{\sqrt{x+h-2}\sqrt{x-2}(\sqrt{x-2} + \sqrt{x+h-2})} $$
$$ = -\frac{8}{\sqrt{x+h-2}\sqrt{x-2}(\sqrt{x-2} + \sqrt{x+h-2})}. $$
When $h$ approaches $0$, the last fraction approaches:
$$  -\frac{4}{(x-2)\sqrt{x-2}} $$
