Derivative of Exponential Function

Theorem
Let $\exp$ be the exponential function.

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

Proof 1
We have:

From one of the definitions of the exponential function,

The right summand converges to zero as $h$ gets arbitrarily small, and so:
 * $\displaystyle \lim_{h \to 0}\frac{\exp h - 1} h = 1$

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.

Proof 2
We use the fact that the exponential function is the inverse of the natural logarithm function:


 * $y = e^x \iff x = \ln y$

Comment
That $D_x \exp x = \exp x$ can be used to define the exponential function.

See Equivalence of Exponential Definitions.