Finding the First Derivative ( 1 question) Using the Definition of a limit: [ Of form $\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$]
Find $f'(x)$ when $x=9$ for $f(x)=\frac{2}{\sqrt{x}}$
I tried simplifying it but got jumbled when trying to multiply by the conjugate, does anyone have a solution?
After some working out i figured the answer.. however is this in the simplest form( the lim.... part not the final answer expressed by the whole numbers):
$\lim_{x\to 9} \frac{-4}{9x-\frac{6+2\sqrt{x}}{3\sqrt{x}}}$
=  -1/27
$\lim_{x\to 9} \frac{-12\sqrt{x}}{(9x)(6+2 \sqrt{x})}$  was the best i got
 A: \begin{equation}f(x)=\frac{2}{\sqrt{x}}\end{equation}
\begin{equation}f'(x)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{2(\sqrt{x}-\sqrt{a})}{\sqrt{ax}(a-x)}: \frac{0}{0}\end{equation}
Use L'Hôpital's rule:
\begin{equation}\frac{2(\sqrt{x}-\sqrt{a})}{\sqrt{ax}(a-x)}=\frac{A(x)}{B(x)}\end{equation}
\begin{equation}A(x)=2(\sqrt{x}-\sqrt{a}),\ B(x)=\sqrt{ax}(a-x)\end{equation}
\begin{equation}\lim_{x\to a}A(x)=0,\ \lim_{x\to a}B(x)=0\end{equation}
\begin{equation}f'(x)=\lim_{x\to a}\frac{2(\sqrt{x}-\sqrt{a})}{\sqrt{ax}(a-x)}=\lim_{x\to a}\frac{A(x)}{B(x)}=\lim_{x\to a}\frac{A'(x)}{B'(x)}=\frac{-1}{x^{\frac{3}{2}}}\end{equation}
\begin{equation}x=9:\ f'(x)=\frac{-1}{27}\end{equation}
A: I post this reply not using De l'Hôpital because deriving an expression that presents an indeterminate form to find the value of its derivative at one point seems quite circular to me. 
$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{2(\sqrt{x}-\sqrt{a})}{\sqrt{ax}(a-x)}=-\lim_{x\to a}\frac{2(\sqrt{x}-\sqrt{a})}{\sqrt{ax}(\sqrt x +\sqrt a)(\sqrt x -\sqrt a)}=\lim_{x\to a} \frac{-2}{\sqrt{ax}(\sqrt x +\sqrt a)}=-a^{-\frac32}$$
