Derivative of $\frac{1}{\sqrt{x+5}}$ 
I'm trying to find the derivative of $\dfrac{1}{\sqrt{x+5}}$
  using $\displaystyle \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}$

So, 
$$\begin{align*}
\lim_{h\to 0} \frac{\dfrac{1}{\sqrt{x+h+5}}-\dfrac{1}{\sqrt{x+5}}}{h} &= \frac{\dfrac{\sqrt{x+5}-\sqrt{x+h+5}}{(\sqrt{x+h+5})(\sqrt{x+5})}}{h}\\\\
&= \frac{\dfrac{x+5-x-h-5}{(\sqrt{x+h+5})(\sqrt{x+5})}}{\dfrac{h}{\sqrt{x+5}}+\dfrac{h}{\sqrt{x+h+5}}}
\end{align*}$$
I do not know if this is correct or not.. please help. I'm stuck.

I would suggest, start it over..
We can introduce a variable, say $u:=x+5$. Then it's
$$\begin{align*} \lim_{h\to 0} \frac1h\cdot \left( \frac1{\sqrt{u+h}} -\frac1{\sqrt{u}} \right) 
&= \lim_{h\to 0} \frac1h\cdot\left(\frac{\sqrt u-\sqrt{u+h}}{\sqrt{u(u+h)}} \right) = \\
&= \lim_{h\to 0} \frac1h\cdot\left(\frac{\sqrt u-\sqrt{u+h}}{\sqrt{u(u+h)}} \cdot \frac{\sqrt{u}+\sqrt{u+h}}{\sqrt{u}+\sqrt{u+h}} \right) = \\
&= \lim_{h\to 0} \frac1h\cdot\left(\frac{u-(u+h)}{\sqrt{u(u+h)}\left(\sqrt{u+h}+\sqrt u\right)}\right) = \\
&=\lim_{h\to 0}\frac{-1}{\sqrt{u(u+h)}\left(\sqrt{u+h}+\sqrt u\right)} =\\
&=\frac{-1}{u\cdot 2\sqrt u}
\end{align*}$$
Finally, rewrite back $u=x+5$.
