The function $f(x) = e^x$ is quite peculiar: it is the only function whose derivative is
itself.

$\displaystyle\frac{d}{dx}(e^x)=e^x$. The
derivative of $e^x$ is $e^x$.

Perhaps $(e^x)'$ is now your favorite derivative.

DO: Find the derivative
of $g(x)=5\cdot e^x$.

What follows is the reasoning behind why $(e^x)'=e^x$.
Please enjoy the following work and video if you are interested in
this unusual function.

We have \begin{eqnarray*} f'(x) & = & \lim_{h \to 0}
\frac{f(x+h)-f(x)}{h} \cr & = & \lim_{h \to 0}
\frac{e^{x+h} - e^x}{h} \cr & = & \lim_{h \to 0}
\frac{e^xe^h - e^x}{h} \cr & = & e^x \cdot\left(\lim_{h
\to 0} \frac{e^h - 1}{h}\right). \end{eqnarray*} The last step was
made possible by the fact that $e^x$ doesn't depend on $h$.
We are left with $$ \lim_{h \to 0} \frac{e^h - 1}{h}. $$ This
limit is 1, which you can guess by exploring this limit
graphically or numerically. Thus, the derivative
of $e^x$ is $e^x$.