# Derivative of Exponential Function/Proof 4

## Theorem

Let $\exp$ be the exponential function.

Then:

- $\map {D_x} {\exp x} = \exp x$

## Proof

This proof assumes the series definition of $\exp$.

That is, let:

- $\displaystyle \exp x = \sum_{k \mathop = 0}^\infty \frac{x^k}{k!}$

From Series of Power over Factorial Converges, the interval of convergence of $\exp$ is the entirety of $\R$.

So we may apply Differentiation of Power Series to $\exp$ for all $x \in \R$.

Thus we have:

\(\displaystyle D_x \exp x\) | \(=\) | \(\displaystyle D_x \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^\infty \frac k {k!} x^{k - 1}\) | Differentiation of Power Series, with $n = 1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^\infty D_x \frac {x^{k - 1} } {\left({k - 1}\right)!}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^\infty D_x \frac {x^k} {k!}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \exp x\) |

Hence the result.

$\blacksquare$