Derivative of Exponential Function

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

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

Proof
Let $$y = \exp x$$. Then:

$$ $$

From Derivative of Exponent at Zero, $$\lim_{x \to 0} \frac {\exp h - 1} {h} = 1$$.

From Combination Theorem for Functions, $$\lim_{h \to 0} \frac {\exp x \left({\exp h - 1}\right)} {h} = \exp x \left({\lim_{x \to 0} \frac {\exp h - 1} {h}}\right) $$.

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:

$$D_x \exp x = \frac 1 {D \ln x} = \frac 1 {1/y} = y = \exp x$$.