Higher Order Derivatives of Laplace Transform

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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:

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


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

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

\(\ds \frac {\d^0} {\d s^0} \laptrans {\map f t}\) \(=\) \(\ds \laptrans {\map f t}\) Definition of Zeroth Derivative
\(\ds \) \(=\) \(\ds \paren {-1}^0 \laptrans {t^0 \, \map f t}\) 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 $n \ge 0$.


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

This is our induction hypothesis.

Induction Step

This is our induction step:

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

The result follows by the Principle of Mathematical Induction.



Example $1$

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

Also see