Derivative of Exponential Function/Proof 4

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Theorem

Let $\exp$ be the exponential function.

Then:

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


Proof

This proof assumes the series definition of $\exp$.

That is, let:

$\displaystyle \exp x = \sum_{k \mathop = 0}^\infty \frac{x^k}{k!}$


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:

\(\displaystyle D_x \exp x\) \(=\) \(\displaystyle D_x \sum_{k \mathop = 0}^\infty \frac {x^k} {k!}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty \frac k {k!} x^{k - 1}\) Differentiation of Power Series, with $n = 1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 1}^\infty D_x \frac {x^{k - 1} } {\left({k - 1}\right)!}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^\infty D_x \frac {x^k} {k!}\)
\(\displaystyle \) \(=\) \(\displaystyle \exp x\)


Hence the result.

$\blacksquare$