Answer
$$
f(x)=e^{x}
$$
Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&=\lim\limits_{h \to 0} \frac{e^{x+h}-e^{x} }{h}\\
&=\lim\limits_{h \to 0} \frac{e^{x}(e^{h}-1) }{h}\\
& \quad\quad\quad\quad[\text { by L'Hospital's Rule }] \\
&=e^{x}
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=e^{2}=7.3891;
$$
$$
f^{\prime}(16)=e^{16}= 8,886,111;
$$
and
$$
f^{\prime}(-3)=e^{-3}= 0.0498.
$$
Work Step by Step
$$
f(x)=e^{x}
$$
Go through the four steps to find $f^{\prime}(x)$
Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$
$$
f(x +h)=e^{x+h}
$$
Step 2
$$
f(x +h) -f(x)=e^{x+h}-e^{x}
$$
Step 3
$$
\frac{f(x +h)- f(x)}{h}= \frac{e^{x+h}-e^{x} }{h}
$$
Step 4 Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&=\lim\limits_{h \to 0} \frac{e^{x+h}-e^{x} }{h}\\
&=\lim\limits_{h \to 0} \frac{e^{x}(e^{h}-1) }{h}\\
& \quad\quad\quad\quad[\text { by L'Hospital's Rule }] \\
&=e^{x}
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=e^{2}=7.3891,
$$
$$
f^{\prime}(16)=e^{16}= 8,886,111
$$
and
$$
f^{\prime}(-3)=e^{-3}= 0.0498
$$
