# f(t)=4e^(-3(t-4)) a) Find L[(df(t))/dt] by differentiating f(t) b) using the theorem for differentiation

Given that $f\left(t\right)=4{e}^{-3\left(t-4\right)}$
a) Find $L\left[\frac{df\left(t\right)}{dt}\right]$ by differentiating f(t) and then using the Laplace transform tables in lecture notes.
b) Find $L\left[\frac{df\left(t\right)}{dt}\right]$ using the theorem for differentiation
c) Repeat a) and b) for the case that $f\left(t\right)=4{e}^{-3\left(t-4\right)}u\left(t-4\right)$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Brittany Patton
Step 1
According to the given information, the given function is:
$f\left(t\right)=4{e}^{-3\left(t-4\right)}$
Step 2
For part (a) it is required to find $L\frac{df\left(t\right)}{dt}$ by differentiating f(t) and then using Laplace transform.
So,
$\frac{df\left(t\right)}{dt}=4{e}^{-3\left(t-4\right)}\left(-3\right)=-12{e}^{-3\left(t-4\right)}$
$L\left(\frac{df}{dt}\right)=L\left(-12{e}^{-3\left(t-4\right)}\right)=L\left(-12{e}^{-3t}{e}^{12}\right)$
$=-12{e}^{12}L\left({e}^{-3t}\right)\left[L\left({e}^{-at}\right)=\frac{1}{s+a}\right]$
$L\left(\frac{df}{dt}\right)=-12{e}^{12}\left(\frac{1}{s+3}\right)$
Step 3
Now, it is required to find $L\left(d\frac{f\left(t\right)}{dt}\right)$ using theorem for differentiation.
$L\left(\frac{df}{dt}\right)=sL\left(f\left(t\right)\right)-f\left(0\right)$
$=s\left(L\left(4{e}^{-3\left(t-4\right)}\right)\right)-4{e}^{-3\left(0-4\right)}$
$=s\left(L\left(4{e}^{-3t}{e}^{12}\right)\right)-4{e}^{12}$
$=4s{e}^{12}L\left({e}^{-3t}\right)-4{e}^{12}$
$=4s{e}^{12}\left(\frac{1}{s+3}\right)-4{e}^{12}$
$=4{e}^{12}\left(\frac{s}{s+3}-1\right)$
$=4{e}^{12}\left(\frac{s-\left(s+3\right)}{s+3}\right)$
$=4{e}^{12}\left(-\frac{3}{s+3}\right)$
$L\left(\frac{df}{dt}\right)=\frac{-12{e}^{12}}{s+3}$
Step 4
Now, for part (c) it is required to repeat part (a) and (b) for the function: