proof of derivative using definition Use the definition to show that the function $f:[0,+\infty)\to \mathbb R$ such that $f(x)=\sqrt{x}$ for all $x\ge 0$ is differentiable at each $x\in (0,+\infty)$.
My solution is
$x_0= (0,\infty +$)
($\lim_{x\rightarrow x_0}\frac{\sqrt{x}-\sqrt{x_0}}{x-x_0}$)
$\frac{\sqrt{x}-\sqrt{x_0}}{x-x_0}*\frac{\sqrt{x}+\sqrt{x_0}}{\sqrt{x}+\sqrt{x_0}}=
\frac{x-x_0}{(x-x_0)(\sqrt{x}+\sqrt{x_0})}=
\frac{1}{\sqrt{x}+\sqrt{x_0}}$ which as $x$ approaches $x_0= \frac{1}{2\sqrt{x_0}}$
I though I was done but I was told the definition to use was $f(x)-f(x_0)=h(x) (x-x_0)$
Any ideas on where i went wrong?

$\lim_{\Delta x \rightarrow 0} 
\frac{\sqrt{x + \Delta{x}}-\sqrt{x}} {\Delta x} 
\frac{\sqrt{x + \Delta{x}} + \sqrt{x}}{\sqrt{x + \Delta{x}} + \sqrt{x}} =
\lim_{\Delta x \rightarrow 0} 
\frac{{x + \Delta{x}}-{x}} {\Delta x {(\sqrt{x + \Delta{x}}} + \sqrt x )} = \frac{1}{2\sqrt{x}}
$
