Derivative of Exponential Function/Proof 4
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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 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.
$\blacksquare$