Derivative of Exponential Function/Proof 5
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Theorem
Let $\exp$ be the exponential function.
Then:
- $\map {\dfrac \d {\d x} } {\exp x} = \exp x$
Proof
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
- $\forall x \in \R : n \ge \ceiling {\size x} \implies \sequence {\dfrac n {n + x} \paren {1 + \dfrac x n}^n}$ 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:
- $\dfrac \d {\d x} \exp x_0 = \exp x_0$
$\blacksquare$