Higher Derivatives of Exponential Function
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Theorem
Let $\exp x$ be the exponential function.
Then:
- $\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
Corollary
Let $c$ be a constant.
Then:
- $\map {\dfrac {\d^n} {\d x^n} } {\map \exp {c x} } = c^n \map \exp {c x}$
Proof
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
- $\map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
Basis for the Induction
$\map P 1$ is true, as this is the case proved in Derivative of Exponential Function:
- $\map {\dfrac \d {\d x} } {\exp x} = \exp x$
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\map {\dfrac {\d^k} {\d x^k} } {\exp x} = \exp x$
Then we need to show:
- $\map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\exp x} = \exp x$
Induction Step
This is our induction step:
\(\ds \map {\dfrac {\d^{k + 1} } {\d x^{k + 1} } } {\exp x}\) | \(=\) | \(\ds \map {\dfrac \d {\d x} } {\dfrac {\d^k} {\d x^k} } {\exp x}\) | Definition of Higher Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\dfrac \d {\d x} } {\exp x}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \exp x\) | Basis for the Induction |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \N_{>0}: \map {\dfrac {\d^n} {\d x^n} } {\exp x} = \exp x$
$\blacksquare$