#### Answer

$\textbf{r}'(t) = \langle 6t^2, \frac{3}{\sqrt t}, \frac{-3}{t^2}\rangle$

#### Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
$\textbf{r}(t) = \langle 2t^3, 6\sqrt t, \frac{3}{t}\rangle$
$\textbf{r}'(t) = \langle 6t^2, \frac{3}{\sqrt t}, \frac{-3}{t^2}\rangle$