Laplace Transform for a difficult function The Laplace Transform I'm having trouble with is:
$$f(t) = 6te^{-9t}\sin(6t)$$
I'm not sure what the protocol is for multiplying t into it.
The Laplace Transform for $f(t) = 6e^{-9t}\sin(6t)$ is $\dfrac 6{(s+9)^2 - 36}$.
Can't figure out how to add in the $t$.
Thanks in advance for your help.

Here is a partial solution:
\begin{align}
\mathcal{L}\{f(t)\}&=\mathcal{L}\{e^{-9t}[6t\sin(6t)]\} \\
&=\mathcal{L}\{6t \sin(6t)\} \vert_{s \to s+9} & \text{First Translation Theorem}\\
&= 6 \mathcal{L}\{t\sin(6t)\} \vert_{s \to s+9} & \text{linearity} \\
&=-6\frac d{ds} \mathcal{L}\{\sin(6t)\} \vert_{s \to s+9} & \text{transform derivative principle} \\
&= -6 \left. \frac d{ds} \frac {6}{s^2+36} \right\vert_{s \to s+9} 
\end{align}
Now evaluate $\frac d{ds} \frac {6}{s^2+36}$, then replace all the $s$ in your derivative with $s+9$.
