Derivative of Exponential Function/Proof 4

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Let $\exp$ be the exponential function.


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


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.