Answer
$$
f(x)=\sqrt {x}
$$
Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&= \lim\limits_{h \to 0}\frac{\sqrt {x+h} -\sqrt {x}}{h}\\
&=\lim\limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\
&=\lim\limits_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\
&=\lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x}) } \\
&=\lim\limits_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}\\
&=\frac{1}{2\sqrt {x}}
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=\frac{1}{2\sqrt {2}}
$$
$$
f^{\prime}(16)=\frac{1}{2\sqrt {16}} =\frac{1}{8}
$$
and
$$
f^{\prime}(-3)=\frac{1}{2\sqrt {-3}} \\
\quad\quad\quad\quad[\text { is not a real number , hence }]
$$
$f^{\prime}(-3)$ does not exist.
Work Step by Step
$$
f(x)=\sqrt {x}
$$
Go through the four steps to find $f^{\prime}(x)$
Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$
$$
f(x +h)=\sqrt {x+h}
$$
Step 2
$$
f(x +h) -f(x)=\sqrt {x+h} -\sqrt {x}
$$
Step 3
$$
\frac{f(x +h)- f(x)}{h}= \frac{\sqrt {x+h} -\sqrt {x}}{h}
$$
Step 4 Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&= \lim\limits_{h \to 0}\frac{\sqrt {x+h} -\sqrt {x}}{h}\\
&=\lim\limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\
&=\lim\limits_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\
&=\lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x}) } \\
&=\lim\limits_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}\\
&=\frac{1}{2\sqrt {x}}
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=\frac{1}{2\sqrt {2}}
$$
$$
f^{\prime}(16)=\frac{1}{2\sqrt {16}} =\frac{1}{8}
$$
and
$$
f^{\prime}(-3)=\frac{1}{2\sqrt {-3}} \\
\quad\quad\quad\quad[\text { is not a real number , hence }]
$$
$f^{\prime}(-3)$ does not exist.
