Derivative of Exponential Function

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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:

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


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.


From Closed Real Interval is Compact in Metric Space, $I$ is compact.

From Chain Rule for Derivatives:

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


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:

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

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

$\blacksquare$


Also see


Sources