# Derive Laplace Transform of the Third Derivative

• Northbysouth
So, if we let ##f(t) = f'''(t)## and the integration by parts formula is ##L\{f(t)\} = \int_0^{\infty} e^{-st}f(t)dt##, then we can write ##L\{f'''(t)\} = L\{f''(t)\} = L\{f'(t)\} = \int_0^{\infty} e^{-st}f'(t)dt##. In summary, the Laplace Transform of the third derivative of f(t) is found by applying integration by parts three times, using the formula ##L\{f(t)\} = \int_0^{\infty

## Homework Statement

Derive he Laplace Transform of the third derivative of f(t).

## The Attempt at a Solution

So, I'm not at all sure how to do this. I think I can start with:

L{f'''(t)} =

But I'm honestly not sure how this works. Any guidance would be appreciated

Northbysouth said:

## Homework Statement

Derive he Laplace Transform of the third derivative of f(t).

## The Attempt at a Solution

So, I'm not at all sure how to do this. I think I can start with:

L{f'''(t)} =

But I'm honestly not sure how this works. Any guidance would be appreciated

First, for a differentiable function F(t), derive the Laplace transform of F'(t) in terms of the transform of F(t)----standard method/material, widely available. Basically, use integration by parts.

Northbysouth said:

## Homework Statement

Derive he Laplace Transform of the third derivative of f(t).

## The Attempt at a Solution

So, I'm not at all sure how to do this. I think I can start with:

L{f'''(t)} =

But I'm honestly not sure how this works. Any guidance would be appreciated

Start with ##L\{f(t)\} = \int_0^{\infty}e^{-st}f(t)dt##. Let ##u = f(t)## and ##dv = e^{-st}## and apply integration by parts. Solve for ##\int_0^{\infty} e^{-st}f'(t)dt##, which is ##L\{f'(t)\}##. There will be an ##f(0)## term in your expression.

This is the standard method for finding the LT of a first derivative. Once you've done this, all you need to do is apply that iteratively (twice) to find the required LT of the third derivative. Note that you don't need to do the integration again, just apply the formula you've derived twice more.