Derivative of Exponential Function/Proof 5

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Let $\exp$ be the exponential function.


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


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 in Metric Space, $I$ is compact.

From Chain Rule for Derivatives:

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


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


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:

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