Derivative of Exponential Function

Theorem
Let $\exp x$ be the exponential function.

Then:
 * $D_x \left({\exp x}\right) = \exp x$

Proof
We have:

From one of the definitions of the exponential function,

The right summand clearly converges to zero as $h$ gets arbitrarily small, and so $\displaystyle \lim_{h \to \infty}\frac{\exp h - 1}{h} = 1$.

From the Multiple Rule for Limits of Functions:
 * $\displaystyle \lim_{h \to 0} \frac {\exp x \left({\exp h - 1}\right)} {h} = \exp x \left({\lim_{x \to 0} \frac {\exp h - 1} {h}}\right)$

The result follows.

Alternative proof
We use the definition of the exponential function as the inverse of the natural logarithm function.

Let $y = \exp x$.

Then from Derivative of an Inverse Function, we have:


 * $\displaystyle D_x \exp x = \frac 1 {D_x \ln x} = \frac 1 {1/y} = y = \exp x$