Answer
$\boldsymbol{r'}(t)=\langle\frac{1}{2\sqrt{t-2}},0,\frac{-2}{t^3}\rangle$.
Work Step by Step
$\boldsymbol{r}(t)=\langle \sqrt{t-2},3,\frac{1}{t^2}\rangle$
In order to compute $\boldsymbol{r'}(t)$ we simply take the derivative of each component with respect to t of $\boldsymbol{r}(t)$.
$\boldsymbol{r'}(t)=\frac{d}{dt}\boldsymbol{r}(t)=\frac{d}{dt}\langle \sqrt{t-2},3,\frac{1}{t^2}\rangle=\langle \frac{d}{dt}\sqrt{t-2},\frac{d}{dt}3,\frac{d}{dt}\frac{1}{t^2}\rangle=\langle\frac{1}{2\sqrt{t-2}},0,\frac{-2}{t^3}\rangle$