Answer
\[
\vec{r}'(t)=\langle \frac{-1}{(1 + t)^2}, \frac{1}{(1 + t)^2}, \frac{t^2+2t}{(1 + t)^2} \rangle
\]
Work Step by Step
\[
\vec{r}(t)=\langle \frac{1}{1 + t}, \frac{t}{1 + t}, \frac{t^2}{1+t} \rangle
\]
In order to compute \(\vec{r}'(t)\) we simply take the derivative of each component with respect to \(t\) of \(\vec{r}(t)\).
\[
\vec{r}'(t)=\langle \frac{-1}{(1 + t)^2}, \frac{1}{(1 + t)^2}, \frac{t^2+2t}{(1 + t)^2} \rangle
\]
