# Derivative of Exponential Function

## Theorem

Let $\exp$ be the exponential function.

Then:

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

### Corollary 1

Let $c \in \R$.

Then:

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

### Corollary 2

Let $a \in \R: a > 0$.

Let $a^x$ be $a$ to the power of $x$.

Then:

$D_x \left({a^x}\right) = a^x \ln a$

## Proof 1

 $\displaystyle \map {D_x} {\exp x}$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\map \exp {x + h} - \exp x} h$ Definition of Derivative $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h$ Exponential of Sum $\displaystyle$ $=$ $\displaystyle \lim_{h \mathop \to 0} \frac {\exp x \paren {\exp h - 1} } h$ $\displaystyle$ $=$ $\displaystyle \exp x \paren {\lim_{h \mathop \to 0} \frac {\exp h - 1} h}$ Multiple Rule for Limits of Functions, as $\exp x$ is constant $\displaystyle$ $=$ $\displaystyle \exp x$ Derivative of Exponential at Zero

$\blacksquare$

## Proof 2

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

$y = e^x \iff x = \ln y$
 $\displaystyle \dfrac {\d x} {\d y}$ $=$ $\displaystyle \dfrac 1 y$ Derivative of Natural Logarithm Function $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\d y} {\d x}$ $=$ $\displaystyle \dfrac 1 {1 / y}$ Derivative of Inverse Function $\displaystyle$ $=$ $\displaystyle y$ $\displaystyle$ $=$ $\displaystyle e^x$

$\blacksquare$

## Proof 3

 $\displaystyle D_x (\ln e^x)$ $=$ $\displaystyle D_x (x)$ 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 $e^x$

$\blacksquare$

## Proof 4

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.

$\blacksquare$

## Proof 5

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

So let:

$\displaystyle \forall n \in \N: \forall x \in \R: f_n \left({x}\right) = \left({1 + \frac x n}\right)^n$

Let $x_0 \in \R$.

Consider $I := \left[{x_0 - 1, \,.\,.\, x_0 + 1}\right]$.

Let:

$N = \left\lceil{\max \left\{ {\left\vert{x_0 - 1}\right\vert, \left\vert{x_0 + 1}\right\vert} \right\} }\right\rceil$

where $\left\lceil{ \cdot }\right\rceil$ denotes the ceiling function.

From Closed Real Interval is Compact, $I$ is compact.

From Chain Rule:

$\displaystyle D_x f_n \left({x}\right) = \frac n {n + x} f_n \left({x}\right)$

### Lemma

$\forall x \in \R : n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil \implies \left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is increasing.

$\Box$

From the lemma:

$\displaystyle \forall x \in I : \left\langle{D_x f_{n + N} \left({x}\right)}\right\rangle$ is increasing

Hence, from Dini's Theorem, $\left\langle{D_x f_{n + N} }\right\rangle$ is uniformly convergent on $I$.

Therefore, for $x \in I$:

 $\displaystyle D_x \exp x$ $=$ $\displaystyle D_x \lim_{n \mathop \to \infty} f_n \left({x}\right)$ $\displaystyle$ $=$ $\displaystyle D_x \lim_{n \mathop \to \infty} f_{n + N} \left({x}\right)$ Tail of Convergent Sequence $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} D_x f_{n + N} \left({x}\right)$ Derivative of Uniformly Convergent Sequence of Differentiable Functions $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \frac n {n + x} f_n \left({x}\right)$ from above $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} f_n \left({x}\right)$ Combination Theorem for Sequences $\displaystyle$ $=$ $\displaystyle \exp x$

In particular:

$D_x \exp x_0 = \exp x_0$

$\blacksquare$