## 1. Intuitive introduction (first derivation)

Let f be a function such that

F(s)=\mathscr{L}\{f(t)\}

By definition, we have:

\displaystyle F(s)=\int_{0}^{\infty}e^{-st}f(t)dt

If we differentiate both sides above, we get

\displaystyle \frac{d}{ds}F(s)=\frac{d}{ds}\int_{0}^{\infty}e^{-st}f(t)dt

i.e,

\displaystyle \frac{d}{ds}F(s)=\int_{0}^{\infty}\frac{\partial}{\partial s}[e^{-st}f(t)]dt

Since f(t) is a function in the variable t, so it is constant with respect to s:

\begin{aligned} \frac{\partial}{\partial s}[e^{-st}f(t)] & = \frac{\partial }{\partial s}(e^{-st})\cdot f(t) \\ & = -te^{-st}f(t) \end{aligned}

Hence,

\displaystyle \frac{d}{ds}F(s)=-\int_{0}^{\infty}e^{-st}tf(t)dt

And as the integral above is the Laplace Transform of the function tf(t), then we may conclude the following:

## Illustrative Example t*sin(2t)

Calculate the Laplace transform of the function t\sin(2t) .

We know that Lapalce transform of sin(at) is given by:

\displaystyle \mathscr{L}\{\sin(at)\}=\frac{a}{s^{2}+a^{2}}

Then, for a=2:

\displaystyle\mathscr{L}\{\sin(2t)\}=\frac{2}{s^{2}+2^{2}}

i.e.,

\displaystyle F(s)=\frac{2}{s^{2}+4}

Therefore,

\displaystyle\mathscr{L}\{t\sin(2t)\}=-\frac{d}{ds}\left(\frac{2}{s^{2}+4}\right)

And by the quotient rule:

\begin{aligned} \mathscr{L}\{t\sin(2t)\} & = -\left(\frac{(2)'\cdot(s^{2}+4)-2\cdot(s^{2}+4)'}{(s^{2}+4)^{2}}\right) \\ & = -\left(\frac{0\cdot(s^{2}+4-2\cdot2s}{(s^{2}+4)^{2}}\right) \\ & = -\frac{-4s}{(s^{2}+4)^{2}} \\ & = \frac{4s}{(s^{2}+4)^{2}} \end{aligned}

## 2. What about the second derivative?

Now, taking the second derivative of F(s):

\begin{aligned} \frac{d^{2}}{ds^{2}}F(s) & = \frac{d}{ds}\left(\frac{d}{ds}F(s)\right) \\ & = \frac{d}{ds}\int_{0}^{\infty}e^{-st}tf(t)dt \end{aligned}

We may proceed as we did before:

\displaystyle\frac{d^{2}}{ds^{2}}F(s)=\int_{0}^{\infty}\frac{\partial}{\partial s}(e^{-st})tf(t)dt

which gives us

\displaystyle\frac{d^{2}}{ds^{2}}F(s)=\int_{0}^{\infty}e^{-st}t^{2}f(t)dt

Hence, we have the following definition:

## Illustrative example t*t*sin(2t)

Calculate the Laplace transform of the function t^2\sin(2t) .

From the last example:

\displaystyle\mathscr{L}\{\sin(2t)\}=\frac{4}{s^{2}+4}=F(s)

and

\displaystyle\frac{d}{ds}\left(\frac{4}{s^{2}+4}\right)=\frac{-4s}{(s^{2}+4)^{2}}

As

\displaystyle\mathscr{L}\{t^{2}\sin(2t)\}=\frac{d^{2}}{ds^{2}}F(s)

to evaluate the Laplace Transform above, we just need to derivate the expression above, i.e.,

\displaystyle\mathscr{L}\{t^{2}\sin(2t)\}=\frac{d}{ds}\left(\frac{-4s}{(s^{2}+4)^{2}}\right)

By the quotient and chain rules

\begin{aligned} \frac{d}{ds}\left(\frac{-4s}{(s^{2}+4)^{2}}\right) & = \frac{(-4s)'(s^{2}+4)^{2}-(-4s)[(s^{2}+4)^{2}]'}{[(s^{2}+4)^{2}]^{2}} \\ & = \frac{-4(s^{2}+4)^{2}+4s\cdot2(s^{2}+4)(s^{2}+4)'}{(s^{2}+4)^{4}} \\ & = \frac{-4(s^{2}+4)^{2}+8s(s^{2}+4)\cdot2s}{(s^{2}+4)^{2}} \\ \end{aligned}

And factoring

\begin{aligned} \frac{d}{ds}\left(\frac{-4s}{(s^{2}+4)^{2}}\right) & = \frac{(s^{2}+4)(-4(s^{2}+4)+8s\cdot2s}{(s^{2}+4)^{4}} \\ & = \frac{-4s^{2}-16+16s^{2}}{(s^{2}+4)^{3}} \\ & = \frac{12s^{2}-16}{(s^{2}+4)^{3}} \end{aligned}

Thus

\displaystyle\mathscr{L}\{t^{2}\sin(2t)\}=\frac{12s^{2}-16}{(s^{2}+4)^{3}}

## 3. Theorem

If we follow the same procedure we used in this section, we have the following

We reached the end of this short lesson about the frequency differentiation property of the Laplace transform. For a full list of Laplace transform properties, check this post!