Finding derivative of $\sqrt{9-x}$ I am trying to find the derivative of $\sqrt{9-x}$ using the definition of a derivative 
$$\lim_{h\to 0} \frac {f(a+h)-f(a)}{h} $$
$$\lim_{h\to 0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h} $$
So to simplify I multiply by the conjugate
$$\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}$$
which gives me 
$$\frac {-2a-h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}$$
I have no idea what to do from here, obviously I can easily get the derivative using other methods but with this one I have no idea how to proceed.

Everything you have done is right except for the last step.
$$\begin{align}
&\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}=\\
&\lim_{h\to0} \frac{9-(a+h)-(9-a)}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{9-a-h-9+a}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{1}{\sqrt{9-(a+h)}+\sqrt{9-a}}
\end{align}$$
The limit is then easy to evaluate.
