Answer
$\displaystyle \frac{d}{dt}[\mathrm{r}(t)\times \mathrm{r}^{\prime}(t)]=\mathrm{r}(t)\times \mathrm{r}^{\prime\prime}(t)$.
(see proof in step-by-step)
Work Step by Step
Applying Theorem 3.5,
$\displaystyle \frac{d}{dt}[\mathrm{u}(t)\times \mathrm{v}(t)]=\mathrm{u}^{\prime}(t)\times \mathrm{v}(t)+\mathrm{u}(t)\times \mathrm{v}^{\prime}(t)$ ,
$\displaystyle \frac{d}{dt}[\mathrm{r}(t)\times \mathrm{r}^{\prime}(t)]=\mathrm{r}^{\prime}(t)\times \mathrm{r}^{\prime}(t)+\mathrm{r}(t)\times \mathrm{r}^{\prime\prime}(t)$
A property of the cross product is that $\mathrm{a}\times \mathrm{a}=0$
(This is because the length of the vector obtained by the cross product $\mathrm{a}\times \mathrm{b}$ depends on the sine of the angle between $\mathrm{a}$ and $\mathrm{b}.$
The angle between a and itself is zero, so then is the sine, so then is the cross product.)
So, since$\quad \mathrm{r}^{\prime}(t)\times \mathrm{r}^{\prime}(t)=0$
we have
$\displaystyle \frac{d}{dt}[\mathrm{r}(t)\times \mathrm{r}^{\prime}(t)]=\mathrm{r}(t)\times \mathrm{r}^{\prime\prime}(t)$.
