$\displaystyle t^3cost-13tsint-13cost$

Thanks

Jason

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- Aug 5th 2009, 08:18 AMDarkhrse99Can I get help finding the derivative?
$\displaystyle t^3cost-13tsint-13cost$

Thanks

Jason - Aug 5th 2009, 08:25 AMAlephZero
- Aug 5th 2009, 09:46 AMDarkhrse99
I came up with $\displaystyle -3t^2sint-13cost+13sint$

Is this correct? - Aug 5th 2009, 09:58 AMFailure
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}$