$\displaystyle t^3cost-13tsint-13cost$
Thanks
Jason
Not quite. You do not seem to have applied the product rule correctly.
I get the following (for the the first two terms you need to apply the product rule)
$\displaystyle \begin{array}{l}
(t^3\cos t-13t\sin t-13\cos t)'\\
\quad =3t^2\cdot\cos t+t^3\cdot(-\sin t)-13\cdot \sin t-13t\cdot\cos t-13(-\sin t)\\
\quad = 3t^2\cos t -t^3\sin t-13t\cos t
\end{array}$