Answer
$\boldsymbol{r'}(t)=\langle2a\sin{at}\cos{at},e^{bt}+bte^{bt},-2c\cos{ct}\sin{ct}\rangle$.
Work Step by Step
$\boldsymbol{r}(t)=\langle \sin^2{at},te^{bt},\cos^2{ct}\rangle$
In order to compute $\boldsymbol{r'}(t)$ we simply take the derivative of each component with respect to t of $\boldsymbol{r}(t)$.
$\boldsymbol{r'}(t)=\frac{d}{dt}\boldsymbol{r}(t)=\frac{d}{dt}\langle \sin^2{at},te^{bt},\cos^2{ct}\rangle=\langle \frac{d}{dt}\sin^2{at},\frac{d}{dt}te^{bt},\frac{d}{dt}\cos^2{ct}\rangle=\langle2a\sin{at}\cos{at},e^{bt}+bte^{bt},-2c\cos{ct}\sin{ct}\rangle$