Can I get help finding the derivative?

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



Thanks
Jason

Quote:

---

Originally Posted by Darkhrse99

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



Thanks
Jason

---

The derivative is found using relatively straight-forward methods (Product Rule and Summing Rule). Can you show your work, so that we can see what you've done?

I came up with $\displaystyle -3t^2sint-13cost+13sint$

Is this correct?

Quote:

---

Originally Posted by Darkhrse99

I came up with $\displaystyle -3t^2sint-13cost+13sint$

Is this correct?

---

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)

$\displaystyle \begin{array}{l}
(t^3\cos t-13t\sin t-13\cos t)'\\
\quad =3t^2\cdot\cos t+t^3\cdot(-\sin t)-13\cdot \sin t-13t\cdot\cos t-13(-\sin t)\\
\quad = 3t^2\cos t -t^3\sin t-13t\cos t
\end{array}$