# Higher Order Derivatives of Laplace Transform

## Theorem

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function on any interval of the form $0 \le t \le A$.

Let $\dfrac {\partial f}{\partial s}$, the partial derivative of $f$ with respect to $s$, exist and be continuous on said intervals.

Let $\laptrans f = F$ denote the Laplace transform of $f$.

Then, everywhere that $\laptrans f$ exists and is $n$ times differentiable:

$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$

## Proof

The proof proceeds by induction on $n$, the order of the derivative of $\laptrans f$.

The case $n = 0$ is verified as follows:

 $\displaystyle \frac {\d^0} {\d s^0} \laptrans {\map f t}$ $=$ $\displaystyle \laptrans {\map f t}$ Definition of Zeroth Derivative $\displaystyle$ $=$ $\displaystyle \paren {-1}^0 \laptrans {t^0 \, \map f t}$ Definition of Zeroth Power

### Basis for the Induction

The case $n = 1$ is demonstrated in Derivative of Laplace Transform:

$\dfrac \d {\d s} \laptrans {\map f t} = -\laptrans {t \, \map f t}$

This is the basis for the induction.

### Induction Hypothesis

Fix $n \in \N$ with $n \ge 0$.

Assume:

$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$

This is our induction hypothesis.

### Induction Step

This is our induction step:

 $\displaystyle \frac {\d^{n + 1} } {\d s^{n + 1} } \laptrans {\map f t}$ $=$ $\displaystyle \frac {\d}{\d s} \frac {\d^n} {\d s^n} \laptrans {\map f t}$ $\displaystyle$ $=$ $\displaystyle \frac {\d}{\d s} \paren {\paren {-1}^n \laptrans {t^n \, \map f t} }$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \paren {-1}^n \paren {-\laptrans {t \times t^n \, \map f t} }$ Derivative of Laplace Transform $\displaystyle$ $=$ $\displaystyle \paren {-1}^{n + 1} \laptrans {t^{n + 1} \, \map f t}$ simplification

The result follows by the Principle of Mathematical Induction.

$\blacksquare$

## Examples

### Example $1$

$\laptrans {t^2 e^{2 t} } = \dfrac 2 {\paren {s - 2}^3}$