# Laplace Transform of Derivatives: Formula, Proof

If the Laplace transform of a function f(t) is known, then the Laplace transform of its derivative f'(t) can be computed. Actually, one can find the Laplace transform of (any finite order) derivatives of f(t). In this article, we will learn how to find the Laplace transform of derivatives and use it to solve an ordinary differential equation as an application.

## First order derivative

Proof:

By the definition of Laplace transforms, we have that

L{f$’$(t)} = $\int_0^\infty e^{-st} f'(t) dt$

Integrating the above by parts, we get that

L{f$’$(t)} = $\Big[ e^{-st} f(t) \Big]_0^\infty$ – $\int_0^\infty (-se^{-st}) f(t) dt$

As e-st f(t)0 when t∞, from above it follows that

L{f$’$(t)} = -f(0) +s $\int_0^\infty e^{-st} f(t) dt$

= -f(0) + s L{f(t)}.

So we have proved that L{f$’$(t)} = -f(0) + s L{f(t)} and this is the Laplace transform formula of first order derivative.

Remark: Note that if f(0)=0, then L{f$’$(t)} = s L{f(t)}, that is, the Laplace transform of the derivative of f(t) is equal to the multiplication of the Laplace transform of f(t) by s.

## Second order derivative

Proof:

From above, we have that

L{f$’$(t)} = -f(0) + s L{f(t)} …(∗)

Replacing f(t) by f$’$(t) in this formula, we get that

L{f$^{”}$(t)} = -f$’$(0) + s L{f$’$(t)}

= -f$’$(0) + s (-f(0) + s L{f(t)}) by (∗)

= s2 L{f(t)} – sf(0) -f$’$(0).

Thus, we have shown that L{f$^{”}$(t)} = s2 L{f(t)} – sf(0) -f$’$(0) and this is the Laplace transform formula of second order derivatives.

Related Topics:

Laplace Transform: Definition, Table, Formulas, Properties

Solved problems of Laplace transforms