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.
Best Answer
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 $$\eqalign{ f'(a)= \lim_{h\rightarrow 0} { -h\over h\bigl( \sqrt{9-(a-h) }+\sqrt{9-a}\ \bigr ) } &=\lim_{h\rightarrow 0} { -1\over \sqrt{9-(a-h) }+\sqrt{9-a} }\cr &= { -1\over \sqrt{9-(a-0) }+\sqrt{9-a} }\cr &= { -1\over \sqrt{9-a }+\sqrt{9-a} }\cr &= { -1\over 2\sqrt{9-a} }. } $$