Answer
$\mathbf{r}'(t)=\langle-e^{-t}, 1-3t^2, \frac{1}{t}\rangle$
Work Step by Step
1) Differentiate $\mathbf{i}$ component of $\mathbf{r}(t)$ using chain rule.
--> $\frac{d(e^{-t}\mathbf{i})}{dt}=(e^{-t})(-1)\mathbf{i}=-e^{-t}\mathbf{i}$
2) Differentiate $\mathbf{j}$ component of $\mathbf{r}(t)$ using power rule
--> $\frac{d(t-t^3\mathbf{j})}{dt}=(1-3t^2)\mathbf{j}$
3) Differentiate $\mathbf{k}$ component of $\mathbf{r}(t)$
--> $\frac{d(ln(t)\mathbf{k})}{dt}=\frac{1}{t}\mathbf{k}$
4) Therefore, $\mathbf{r}'(t)=-e^{-t}\mathbf{i}+(1-3t^2)\mathbf{j}+\frac{1}{t}\mathbf{k}=\langle-e^{-t}, 1-3t^2, \frac{1}{t}\rangle$
