### 2 The Laplace transform of a derivative

Here we consider not a causal function $f\left(t\right)$ directly but its derivatives $\frac{df}{dt}$ , $\frac{{d}^{2}f}{d{t}^{2}}$ , …(which are also causal.) The Laplace transform of derivatives will be invaluable when we apply the Laplace transform to the solution of constant coefficient ordinary differential equations.

If $\mathsc{L}\left\{f\left(t\right)\right\}$ is $F\left(s\right)$ then we shall seek an expression for $\mathsc{L}\left\{\frac{df}{dt}\right\}$ in terms of the function $F\left(s\right)$ .

Now, by the definition of the Laplace transform

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{\frac{df}{dt}\right\}={\int }_{0}^{\infty }{\text{e}}^{-st}\frac{df}{dt}\phantom{\rule{0.3em}{0ex}}dt$

This integral can be simplified using integration by parts:

$\begin{array}{rcll}{\int }_{0}^{\infty }{\text{e}}^{-st}\frac{df}{dt}\phantom{\rule{0.3em}{0ex}}dt& =& {\left[{\text{e}}^{-st}f\left(t\right)\right]}_{0}^{\infty }-{\int }_{0}^{\infty }\left(-s\right){\text{e}}^{-st}f\left(t\right)\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& -f\left(0\right)+s{\int }_{0}^{\infty }{\text{e}}^{-st}f\left(t\right)\phantom{\rule{0.3em}{0ex}}dt& \text{}\end{array}$

(As usual, we assume that contributions arising from the upper limit, $t=\infty$ , are zero.) The integral on the right-hand side is precisely the Laplace transform of $f\left(t\right)$ which we naturally replace by $F\left(s\right)$ . Thus

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{\frac{df}{dt}\right\}=-f\left(0\right)+sF\left(s\right)$

As an example, we know that if $f\left(t\right)=sint\phantom{\rule{1em}{0ex}}u\left(t\right)$ then

and so, according to the result just obtained,

$\begin{array}{rcll}\mathsc{L}\left\{\frac{df}{dt}\right\}=\mathsc{L}\left\{cost\phantom{\rule{1em}{0ex}}u\left(t\right)\right\}& =& -f\left(0\right)+sF\left(s\right)& \text{}\\ & =& 0+s\left(\frac{1}{{s}^{2}+1}\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& \frac{s}{{s}^{2}+1}& \text{}\end{array}$

a result we know to be true.

We can find the Laplace transform of the second derivative in a similar way to find:

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{\frac{{d}^{2}f}{d{t}^{2}}\right\}=-{f}^{\prime }\left(0\right)-sf\left(0\right)+{s}^{2}F\left(s\right)$

(The reader might wish to derive this result.) Here ${f}^{\prime }\left(0\right)$ is the derivative of $f\left(t\right)$ evaluated at $t=0$ .

##### Key Point 10

Laplace Transforms of Derivatives

If $\mathsc{L}\left\{f\left(t\right)\right\}=F\left(s\right)$ then

$\begin{array}{rcll}\mathsc{L}\left\{\frac{df}{dt}\right\}& =& -f\left(0\right)+sF\left(s\right)& \text{}\\ \mathsc{L}\left\{\frac{{d}^{2}f}{d{t}^{2}}\right\}& =& -{f}^{\prime }\left(0\right)-sf\left(0\right)+{s}^{2}F\left(s\right)& \text{}\end{array}$

If with initial conditions

$f\left(0\right)=1,\phantom{\rule{1em}{0ex}}{f}^{\prime }\left(0\right)=0$ , find the explicit expression for $F\left(s\right)$ .

Begin by finding :

$\begin{array}{rcll}\mathsc{L}\left\{3t\right\}& =& 3∕{s}^{2}& \text{}\\ \mathsc{L}\left\{\frac{df}{dt}\right\}& =& -f\left(0\right)+sF\left(s\right)=-1+sF\left(s\right)& \text{}\\ \mathsc{L}\left\{\frac{{d}^{2}f}{d{t}^{2}}\right\}& =& -{f}^{\prime }\left(0\right)-sf\left(0\right)+{s}^{2}F\left(s\right)=-s+{s}^{2}F\left(s\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Now complete the calculation to find $F\left(s\right)$ :

You should find $\phantom{\rule{1em}{0ex}}F\left(s\right)=\frac{{s}^{3}-{s}^{2}+3}{{s}^{3}\left(s-1\right)}$ since, using the transforms we have found:

leading to $F\left(s\right)=\frac{{s}^{3}-{s}^{2}+3}{{s}^{3}\left(s-1\right)}$

##### Exercises
1. Find the Laplace transforms of
1. ${t}^{3}{\text{e}}^{-2t}u\left(t\right)$
2. ${\text{e}}^{t}sinh3t.u\left(t\right)$
3. $sin\left(t-3\right).u\left(t-3\right)$
2. If $F\left(s\right)=\mathsc{L}\left\{f\left(t\right)\right\}$ find expressions for $F\left(s\right)$ if
1. $\frac{{d}^{2}y}{d{t}^{2}}-3\frac{dy}{dt}+4y=sint\phantom{\rule{2em}{0ex}}y\left(0\right)=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{y}^{\prime }\left(0\right)=0$
2. $7\frac{dy}{dt}-6y=3u\left(t\right)\phantom{\rule{2em}{0ex}}y\left(0\right)=0,$
3. Find the inverse Laplace transforms of
1. $\frac{6}{{\left(s+3\right)}^{4}}$
2. $\frac{15}{{s}^{2}-2s+10}$
3. $\frac{3{s}^{2}+11s+14}{{s}^{3}+2{s}^{2}-11s-52}$
4. $\frac{{\text{e}}^{-3s}}{{s}^{4}}$
5. $\frac{{\text{e}}^{-2s-2}\left(s+1\right)}{{s}^{2}+2s+5}$
1. $\frac{6}{{\left(s+2\right)}^{4}}$
2. $\frac{3}{{\left(s-1\right)}^{2}-9}$
3. $\frac{{\text{e}}^{-3s}}{{s}^{2}+1}$
1. $\frac{{s}^{3}-3{s}^{2}+s-2}{\left({s}^{2}+1\right)\left({s}^{2}-3s+4\right)}$
2. $\frac{3}{s\left(7s-6\right)}$
1. ${\text{e}}^{-3t}{t}^{3}u\left(t\right)$
2. $5{\text{e}}^{t}sin3t.u\left(t\right)$
3. $\left(2{\text{e}}^{4t}+{\text{e}}^{-3t}cos2t\right)u\left(t\right)$
4. $\frac{1}{6}{\left(t-3\right)}^{3}u\left(t-3\right)$
5. ${\text{e}}^{-t}cos2\left(t-2\right).u\left(t-2\right)$