# Thread: [SOLVED] Derivative properites of LaPlace Transforms

1. ## [SOLVED] Derivative properites of LaPlace Transforms

Given that $g(t)$ has the Laplace transform of $G(s)$, what is the Laplace transform of H(s):

$h(t)=e^{-t}g'(3t)$

I know that $g(t)$ has to satisfy the Derivatives proptery theorem: $f'(t)=s F(s)-f(0)$, but im not really sure what to do. As always step by step instructions would be amazing

2. Originally Posted by snaes
Given that $g(t)$ has the Laplace transform of $G(s)$, what is the Laplace transform of H(s):

$h(t)=e^{-t}g'(3t)$

I know that $g(t)$ has to satisfy the Derivatives proptery theorem: $f'(t)=s F(s)-f(0)$, but im not really sure what to do. As always step by step instructions would be amazing
Note that from the definition it easily follows that $LT[f(at)] = \frac{1}{a} F\left(\frac{s}{a}\right)$ where $F(s) = LT[f(t)]$.

3. Would these be correct?

1)
$h(t)=g'(3t)$
$H(s)=s*LT[g(3t)]-g(0)$
$H(s)=s*\dfrac{1}{3}G(\dfrac{s}{3})-g(0)$

2) Should be the same only only I use $e^{at}f(t) = F(s-a)$
$h(t)=e^{3t}g'(3t)$
$H(s)=(s-3)*\dfrac{1}{3}G(\dfrac{s-3}{3})-g(0)$

3) Now I use $t^{n}f(t)=(-1)^{n}\dfrac{d^{n}}{ds^{n}}(F(s))$
$h(t)=t^{2}e^{3t}g'(3t)$
$H(s)=t^{2}((s-3)*\dfrac{1}{3}G(\dfrac{s-3}{3})-g(0))$
$H(s)=(-1)^{2}\dfrac{d^{2}}{ds^{2}}[(s-3)*\dfrac{1}{3}G(\dfrac{s-3}{3})-g(0)]$
$H(s)=\dfrac{d^{2}}{ds^{2}}[(s/3-1)*G(\dfrac{s-3}{3})$

I think this is right but im not sure...

NVM I GOT IT FIGURED OUT