---
tag: 0K090GVHVN
kind: statement
role: proposition
name: Laplace transform of a derivative
msc: [44A10]
uses: [05XT5AVJBK]
sources:
  - url: https://en.wikipedia.org/wiki/Laplace_transform
    locator: Transform of a function's derivative
  - url: https://en.wikipedia.org/wiki/Laplace_transform
    locator: Properties and theorems
---

## Statement

Let $f : [0, \infty) \to \mathbb{C}$ be continuously differentiable, and let $M, a$ be real constants with $|f(t)| \le M e^{a t}$ for all $t \ge 0$.
Let $F$ be the [[Laplace transform|05XT5AVJBK]] of $f$.
Then for each $s \in \mathbb{C}$ with $\operatorname{Re} s > a$, the Laplace transform of $f'$ converges at $s$ and
$$\mathcal{L}\{f'\}(s) = s F(s) - f(0).$$

## Remarks

Proof: integration by parts gives $\int_0^T f'(t) e^{-st}\, dt = f(T) e^{-sT} - f(0) + s \int_0^T f(t) e^{-st}\, dt$. The term $f(T) e^{-sT}$ tends to $0$ for $\operatorname{Re} s > a$.
Applied twice, under the same hypotheses on $f$ and $f'$: $\mathcal{L}\{f''\}(s) = s^2 F(s) - s f(0) - f'(0)$.
This rule is the basis of [[solving initial value problems with the Laplace transform|0K3HFZJK5Q]]. The initial values enter the transformed equation directly.

## Explanations

### L1 human

Differentiation in time becomes multiplication by $s$ after the transform, with a correction by the initial value.
So a differential equation becomes an algebraic equation for $F$.