Answer
$\mathrm{u}^{\prime}(t)=\mathrm{r}(t)\cdot[\mathrm{r}^{\prime}(t)\times \mathrm{r}^{\prime\prime\prime}(t)]$
(proof in the step-by-step section)
Work Step by Step
Given $\mathrm{u}(t)=\mathrm{r}(t)\cdot[\mathrm{r}^{\prime}(t)\times \mathrm{r}^{\prime\prime}(t)]$,
we apply formula 4 of Th.3:
$\displaystyle \mathrm{u}^{\prime}(t)=\mathrm{r}^{\prime}(t)\cdot[\mathrm{r}^{\prime}(t)\times \mathrm{r}^{\prime\prime}(t)]+\mathrm{r}(t)\cdot\frac{d}{dt}[\mathrm{r}^{\prime}(t)\times \mathrm{r}^{\prime\prime}(t)]$
...vectors $\mathrm{a}$ and $\mathrm{a}\times \mathrm{b}$ are perpendicular, so the first term is a scalar product of perpendicular vectors ...equals 0
... for the second term, apply formula 5 of Th.3:
$=0+\mathrm{r}(t)\cdot[\mathrm{r}^{\prime\prime}(t)\times \mathrm{r}^{\prime\prime}(t)+\mathrm{r}^{\prime}(t)\times \mathrm{r}^{\prime\prime\prime}(t)]$
...For any $\mathrm{a}$, the cross product $\mathrm{a}\times \mathrm{a}=0$, so the first term in the brackets is 0....
$=\mathrm{r}(t)\cdot[\mathrm{r}^{\prime}(t)\times \mathrm{r}^{\prime\prime\prime}(t)]$
