Answer
$\displaystyle \frac{d}{dt}|\mathrm{r}(t)|=\frac{1}{|\mathrm{r}(t)|}\mathrm{r}(t)\cdot \mathrm{r}^{\prime}(t)$
(proof in step-by-step)
Work Step by Step
Use the hint...
$\displaystyle \frac{d}{dt}|\mathrm{r}(t)|=\frac{d}{dt}[\mathrm{r}(t)\cdot \mathrm{r}(t)]^{1/2}$
... use Th.3.4 and 6 (product and chain rule)
$=\displaystyle \frac{1}{2}[\mathrm{r}(t)\cdot \mathrm{r}(t)]^{-1/2}[\mathrm{r}^{\prime}(t)\cdot \mathrm{r}(t$+$\mathrm{r}(t)\cdot \mathrm{r}^{\prime}(t)]$
$=\displaystyle \frac{1}{2}[\mathrm{r}(t)\cdot \mathrm{r}(t)]^{-1/2}[2\mathrm{r}(t)\cdot \mathrm{r}^{\prime}(t)]$
...and the hint, again
$=\displaystyle \frac{1}{|\mathrm{r}(t)|}\mathrm{r}(t)\cdot \mathrm{r}^{\prime}(t)$