Derivative of Exponential Function

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Let $\exp$ be the exponential function.


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

Corollary 1

Let $c \in \R$.


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

Corollary 2

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

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


$\map {D_x} {a^x} = a^x \ln a$

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


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
\(\displaystyle \) \(=\) \(\displaystyle y\)
\(\displaystyle \) \(=\) \(\displaystyle e^x\)


Proof 3

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


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.


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


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

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

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

From Chain Rule for Derivatives:

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


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


From the lemma:

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

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

Therefore, for $x \in I$:

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

In particular:

$D_x \exp x_0 = \exp x_0$


Also see