Answer
$\displaystyle \mathrm{r}^{\prime}(t)=(2te^{t^{2}})\mathrm{i}\ \ +\ \ \dfrac{3}{1+3t}\mathrm{k}$
Work Step by Step
Parametric equations: $\mathrm{r}(t):\quad\left\{\begin{array}{ll}
x=e^{t^{2}} & \\
y=-1 & \\
y=\ln(1+3t) &
\end{array}\right.$
Differentiate $(\displaystyle \frac{d}{dt})$ each component function.
x, z: chain rule
y: constant
$\mathrm{r}^{\prime}(t):\quad\left\{\begin{array}{ll}
x=e^{t^{2}}\cdot 2t & \\ \\
y=0\\ & \\
z=\dfrac{1}{(1+3t)}\cdot 3 &
\end{array}\right. $
$\displaystyle \mathrm{r}^{\prime}(t)=2te^{t^{2}}\mathrm{i}+\frac{3}{1+3t}\mathrm{k}$
