Answer
$\textbf{r}'(t) = \langle \frac{-1}{(t+1)^2}, \frac{1}{1+t^2}, \frac{1}{t+1}\rangle$
Work Step by Step
$\textbf{r}(t) = \langle f(t), g(t), h(t)\rangle$
$\textbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle$
$\textbf{r}(t) = \langle (t+1)^{-1}, tan^{-1} t, ln(t+1)\rangle$
Chain Rule:
$\textbf{r}'(t) = \langle \frac{-1}{(t+1)^2}, \frac{1}{1+t^2}, \frac{1}{t+1}\rangle$