# Derivative of Exponential Function

## Theorem

Let $\exp$ be the exponential function.

Then:

$\map {\dfrac \d {\d x} } {\exp x} = \exp x$

### Corollary 1

Let $a \in \R$.

Then:

$\map {\dfrac \d {\d x} } {\map \exp {a x} } = a \map \exp {a x}$

### Corollary 2

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

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

Then:

$\map {\dfrac \d {\d x} } {a^x} = a^x \ln a$

### Corollary 3

$\map {\dfrac \d {\d x} } {\map \exp {-x} } = -\map \exp {-x}$

## Proof 1

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

$\blacksquare$

## Proof 2

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

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

$\blacksquare$

## Proof 3

 $\ds \map {\frac \d {\d x} } {\ln e^x}$ $=$ $\ds \map {\frac \d {\d x} } x$ Exponential of Natural Logarithm $\ds \leadsto \ \$ $\ds \frac 1 {e^x} \map {\frac \d {\d x} } {e^x}$ $=$ $\ds 1$ Chain Rule for Derivatives, Derivative of Natural Logarithm Function, Derivative of Identity Function $\ds \leadsto \ \$ $\ds \map {\frac \d {\d x} } {e^x}$ $=$ $\ds e^x$ multiply both sides by $e^x$

$\blacksquare$

## Proof 4

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

That is, let:

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

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:

 $\ds \frac \d {\d x} \exp x$ $=$ $\ds \frac \d {\d x} \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^\infty \frac k {k!} x^{k - 1}$ Differentiation of Power Series, with $n = 1$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^\infty \frac {x^{k - 1} } {\paren {k - 1}!}$ $\ds$ $=$ $\ds \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}$ $\ds$ $=$ $\ds \exp x$

Hence the result.

$\blacksquare$

## Proof 5

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

So let:

$\forall n \in \N: \forall x \in \R: \map {f_n} x = \paren {1 + \dfrac x n}^n$

Let $x_0 \in \R$.

Consider $I := \closedint {x_0 - 1} {x_0 + 1}$.

Let:

$N = \ceiling {\max \set {\size {x_0 - 1}, \size {x_0 + 1} } }$

where $\ceiling {\, \cdot \,}$ denotes the ceiling function.

$\dfrac \d {\d x} \map {f_n} x = \dfrac n {n + x} \map {f_n} x$

### Lemma

$\forall x \in \R : n \ge \ceiling {\size x} \implies \sequence {\dfrac n {n + x} \paren {1 + \dfrac x n}^n}$ is increasing.

$\Box$

From the lemma:

$\forall x \in I: \sequence {\dfrac \d {\d x} \map {f_{n + N} } x}$ is increasing

Hence, from Dini's Theorem, $\sequence {\dfrac \d {\d x} f_{n + N} }$ is uniformly convergent on $I$.

Therefore, for $x \in I$:

 $\ds \frac \d {\d x} \exp x$ $=$ $\ds \frac \d {\d x} \lim_{n \mathop \to \infty} \map {f_n} x$ $\ds$ $=$ $\ds \frac \d {\d x} \lim_{n \mathop \to \infty} \map {f_{n + N} } x$ Tail of Convergent Sequence $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \frac \d {\d x} \map {f_{n + N} } x$ Derivative of Uniformly Convergent Sequence of Differentiable Functions $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \frac n {n + x} \map {f_n} x$ from above $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \map {f_n} x$ Combination Theorem for Sequences $\ds$ $=$ $\ds \exp x$

In particular:

$\dfrac \d {\d x} \exp x_0 = \exp x_0$

$\blacksquare$