Derivative of Exponential Function/Proof 5

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Let $\exp$ be the exponential function.


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


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


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

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

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

From Chain Rule:

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


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


From the lemma:

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

$D_x \exp x_0 = \exp x_0$