Higher Derivatives of Exponential Function

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Theorem

Let $\exp x$ be the exponential function.

Let $c$ be a constant.


Then:

$D^n_x \left({\exp x}\right) = \exp x$


Corollary

$D^n_x \left({\exp \left({c x}\right)}\right) = c^n \exp \left({c x}\right)$


Proof

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

$D^n_x \left({\exp x}\right) = \exp x$


Basis for the Induction

$P(1)$ is true, as this is the case proved in Derivative of Exponential Function:

$D_x \left({\exp x}\right) = \exp x$

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.


So this is our induction hypothesis:

$D^k_x \left({\exp x}\right) = \exp x$


Then we need to show:

$D^{k+1}_x \left({\exp x}\right) = \exp x$


Induction Step

This is our induction step:

\(\displaystyle D^{k+1}_x \left({\exp x}\right)\) \(=\) \(\displaystyle D_x \left({D^k_x \left({\exp x}\right)}\right)\) Definition of Higher Derivatives
\(\displaystyle \) \(=\) \(\displaystyle D_x \left({\exp x}\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \exp x\) Basis for the Induction

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N^*: D^n_x \left({\exp x}\right) = \exp x$

$\blacksquare$


Proof of Corollary

This follows directly from Derivatives of Function of $a x + b$:

$D^n_x \left({f \left({a x + b}\right)}\right) = a^n D^n_{z} \left({f \left({z}\right)}\right)$

where $z = a x + b$.


Here we set $a = c$ and $b = 0$ so that:

$D^n_x \left({f \left({c x}\right)}\right) = c^n D^n_{z} \left({f \left({z}\right)}\right)$

where $z = c x$.


Then from the main result:

$D^n_z \left({\exp \left({z}\right)}\right) = \exp z$


Hence the result.

$\blacksquare$