## Calculus: Early Transcendentals 8th Edition

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