Let $\mathbf{c}_1(t) = (e^{-3t}, \sin(t), -3t^3)$, and $\mathbf{c}_2(t) = (e^{3t}, \cos(5t), -3t^3)$

$\displaystyle \frac{d}{dt}\left[ \mathbf{c}_1(t) \cdot \mathbf{c}_2(t)\right] =$
$\displaystyle \frac{d}{dt}\left[ \mathbf{c}_1(t) \times \mathbf{c}_2(t)\right] =$ $\mathbf{i}\ +$
$\hspace{1.15in}$ $\mathbf{j}\ +$
$\hspace{1.15in}$ $\mathbf{k}$