# Derivative of Exponential Function/Complex

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## Theorem

The complex exponential function is its own derivative.

That is:

- $\map {D_z} {\exp z} = \exp z$

## Proof from Sequence Definition

Take the definition of $\exp$ to be the limit of the sequence $\sequence {E_n}$ defined by:

- $\ds \map {E_n} z = \paren {1 + \dfrac z n}^n$

Then $\left \langle{E_n}\right \rangle$ is uniformly convergent on compact subsets of $\C$.

Further, $\C$ is an open, connected subset of $\C$.

So the hypotheses of Derivative of Sequence of Holomorphic Functions are satisfied.

Hence:

\(\ds \map {D_z} {\exp z}\) | \(=\) | \(\ds \map {D_z} {\paren {1 + \dfrac z n}^n}\) | Definition of Complex Exponential Function | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \map {D_z} {\paren {1 + \dfrac z n}^n}\) | Derivative of Sequence of Holomorphic Functions | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {n \paren {1 + \dfrac z n}^{n - 1} \times \frac 1 n}\) | Chain Rule | |||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac z n}^{n - 1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \paren {\paren {1 + \dfrac z n}^n \times \frac n {n + z} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {\lim_{n \mathop \to \infty} \paren {1 + \dfrac z n}^n} \times \paren {\lim_{n \mathop \to \infty} \frac n {n + z} }\) | Complex Derivative of Product is Product of Complex Derivative | |||||||||||

\(\ds \) | \(=\) | \(\ds \exp z \times 1\) | Definition of Complex Exponential Function |

Hence the result.

$\blacksquare$