Answer
$\boldsymbol{r'}(t)=\langle-e^{-t},1-3t^2,\frac{1}{t}\rangle$.
Work Step by Step
$\boldsymbol{r}(t)=\langle e^{-t},t-t^3,\ln{t}\rangle$
In order to compute $\boldsymbol{r'}(t)$ we simply take the derivative of each component with respect to t of $\boldsymbol{r}(t)$.
$\boldsymbol{r'}(t)=\frac{d}{dt}\boldsymbol{r}(t)=\frac{d}{dt}\langle e^{-t},t-t^3,\ln{t}\rangle=\langle \frac{d}{dt}e^{-t},\frac{d}{dt}(t-t^3),\frac{d}{dt}\ln{t}\rangle=\langle-e^{-t},1-3t^2,\frac{1}{t}\rangle$.
