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.
 A: You made a mistake when doing the multiplication upstairs:
When multiplying
$$
\Bigl( \color{maroon}{\sqrt{9-(a+h)} }- \color{darkgreen}{\sqrt {9-a}}\ \Bigr)\Bigl(\color{maroon}{\sqrt{9-(a+h)} }+\color{darkgreen}{ \sqrt {9-a}}\ \Bigr),
$$
you are using the rule
$$
(\color{maroon}a-\color{darkgreen}b)(\color{maroon}a+\color{darkgreen}b)
=\color{maroon}a^2-\color{darkgreen}b^2
$$
So you obtain
$$
\Bigl(\color{maroon}{\sqrt{9-(a+h)}}\ \Bigr)^2 - \Bigl(\color{darkgreen}{\sqrt {9-a}}\ \Bigr)^2= \bigl(9-(a+h)\bigr) - (9-a) = \color{teal}9\color{purple}{-a}-h\color{teal}{-9}+\color{purple} a= -h.
$$

Then, to find your derivative, you have to compute
$$\eqalign{
f'(a)=
\lim_{h\rightarrow 0} { -h\over h\bigl(  \sqrt{9-(a-h) }+\sqrt{9-a}\ \bigr ) }
&=\lim_{h\rightarrow 0} { -1\over   \sqrt{9-(a-h) }+\sqrt{9-a}   }\cr
&=  { -1\over   \sqrt{9-(a-0) }+\sqrt{9-a}   }\cr
&=  { -1\over   \sqrt{9-a }+\sqrt{9-a}   }\cr
&=  { -1\over  2\sqrt{9-a}   }.
}
$$
A: 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.
A: As other answers well say, you have an error while multiplicating. And also you forgot to put the limit in your last equation.
\begin{align}
f'(x)&=\lim\limits_{h\to0}\left(-\frac{h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}\right)\\
&=\lim\limits_{h\to0}\left(-\frac{1}{\sqrt{9-(a+h)}+\sqrt{9-a}}\right)\\
&=-\frac{1}{\sqrt{9-x}+\sqrt{9-x}}\\
&=-\frac{1}{2\sqrt{9-x}}
\end{align}
But this can be done much easily by this way:
\begin{align}
f(x)&=\sqrt{9-x}\\
&=(9-x)^{1/2}\\
f'(x)&=-\frac{1}{2}(9-x)^{-1/2}\\
&=-\frac{1}{2\sqrt{9-x}}
\end{align}
