#### Answer

$\textbf{r}'(1) = \langle 8, 9, -10\rangle$

#### Work Step by Step

$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
$\textbf{r}(t) = \langle 2t^4,6t^{3/2},\frac{10}{t}\rangle$
$\textbf{r}'(t) = \langle 8t^3, 9t^{1/2}, \frac{-10}{t^2}\rangle$
$\textbf{r}'(1) = \langle 8(1)^3, 9(1)^{1/2}, \frac{-10}{(1)^2}\rangle = \langle 8, 9, -10\rangle$