Answer
$\frac{d}{{d{\rm{\theta }}}}{\bf{r}}\left( {\cos \theta } \right) = \left( { - \sin \theta , - 2\sin \theta , - \sin 2\theta } \right)$
Work Step by Step
We have ${\bf{r}}\left( s \right) = \left( {s,2s,{s^2}} \right)$. So, ${\bf{r}}\left( {\cos \theta } \right) = \left( {\cos \theta ,2\cos \theta ,{{\cos }^2}\theta } \right)$. Thus,
$\frac{d}{{d{\rm{\theta }}}}{\bf{r}}\left( {\cos \theta } \right) = \left( { - \sin \theta , - 2\sin \theta , - 2\cos \theta \sin \theta } \right)$
$\frac{d}{{d{\rm{\theta }}}}{\bf{r}}\left( {\cos \theta } \right) = \left( { - \sin \theta , - 2\sin \theta , - \sin 2\theta } \right)$
