Derivative of Exponential Function

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Theorem

Let $\exp$ be the exponential function.

Then:

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


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

\(\displaystyle D_x \left({\exp x}\right)\) \(=\) \(\displaystyle \lim_{h \to 0} \frac {\exp \left({x+h}\right) - \exp x} h\) $\quad$ Definition of Derivative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \to 0} \frac {\exp x \cdot \exp h - \exp x} h\) $\quad$ Exponent of Sum $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{h \to 0} \frac {\exp x \left({\exp h - 1}\right)} h\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \exp x \left({\lim_{h \to 0} \frac {\exp h - 1} h}\right)\) $\quad$ Multiple Rule for Limits of Functions, as $\exp x$ is constant $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \exp x\) $\quad$ Derivative of Exponential at Zero $\quad$

$\blacksquare$


Proof 2

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

$y = e^x \iff x = \ln y$
\(\displaystyle \dfrac {\mathrm d x} {\mathrm d y}\) \(=\) \(\displaystyle \dfrac 1 y\) $\quad$ Derivative of Natural Logarithm Function $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \dfrac {\mathrm d y} {\mathrm d x}\) \(=\) \(\displaystyle \dfrac {1} {1 / y}\) $\quad$ Derivative of Inverse Function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e^x\) $\quad$ $\quad$

$\blacksquare$


Proof 3

\(\displaystyle D_x (\ln e^x)\) \(=\) \(\displaystyle D_x (x)\) $\quad$ Exponential of Natural Logarithm $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \frac{1}{e^x}D_x (e^x)\) \(=\) \(\displaystyle 1\) $\quad$ Chain rule, Derivatives of Natural Log and Identity functions. $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle D_x (e^x)\) \(=\) \(\displaystyle e^x\) $\quad$ multiply both sides by $e^x$ $\quad$

$\blacksquare$


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:

\(\displaystyle D_x \exp x\) \(=\) \(\displaystyle D_x \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty \frac k {k!} x^{k - 1}\) $\quad$ Differentiation of Power Series, with $n = 1$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty D_x \frac {x^{k - 1} } {\left({k - 1}\right)!}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^\infty D_x \frac {x^k} {k!}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \exp x\) $\quad$ $\quad$


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 $x \in I$:

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


In particular:

$D_x \exp x_0 = \exp x_0$

$\blacksquare$


Also see


Sources