Differentiate $f(x)=\sqrt{x}$ and determine the domain of $f^\prime$.

Solution The domain of $f$ is $\{x|x ≥ 0\}$. To find the derivative of $f$, we use form (3). Then $$f^\prime ( x) =\lim\limits_{h\rightarrow 0}\dfrac{f( x+h) -f( x) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{ x+h}-\sqrt{x}}{h}$$

We rationalize the numerator to find the limit. $$\begin{eqnarray*} f^\prime ( x) &=&\lim\limits_{h\rightarrow 0}\left[ \dfrac{\sqrt{ x+h}-\sqrt{x}}{h}\cdot \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x} } \right] =\lim\limits_{h\rightarrow 0}\dfrac{( x+h) -x}{ h( \sqrt{x+h}+\sqrt{x}) } \\[5pt] &=&\lim\limits_{h\rightarrow 0}\dfrac{h}{h( \sqrt{x+h}+\sqrt{x}) } =\lim\limits_{h\rightarrow 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}}=\dfrac{1}{2\sqrt{ x}} \end{eqnarray*}$$

The limit does not exist when $x=0.$ But for all other $x$ in the domain of $f$, the limit does exist. So, the domain of the derivative function $f^\prime (x) =\dfrac{1}{2\sqrt{x}}$ is $\{ x|x>0\}$.