# Derivative of Exponential Function/Proof 3

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## Theorem

Let $\exp$ be the exponential function.

Then:

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

## Proof

\(\displaystyle D_x (\ln e^x)\) | \(=\) | \(\displaystyle D_x (x)\) | Exponential of Natural Logarithm | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \frac{1}{e^x}D_x (e^x)\) | \(=\) | \(\displaystyle 1\) | Chain rule, Derivatives of Natural Log and Identity functions. | |||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle D_x (e^x)\) | \(=\) | \(\displaystyle e^x\) | multiply both sides by $e^x$ |

$\blacksquare$

## Sources

- For a video presentation of the contents of this page, visit the Khan Academy.