Answer
$f^{\prime}(0)=0+2=2$
Work Step by Step
We find:
$r^{\prime}(t)=3 r_{1}^{\prime}(t)+2 r_{2}^{\prime}(t)$
$r^{\prime}(0)=\langle 3,0,3\rangle+\langle 8,0,4\rangle=\langle 11,0,7\rangle$
(b) $r^{\prime}(t)=\frac{1}{t+1} r_{1}(t)+(\ln (t+1)) r_{1}^{\prime}(t)$
$\mathrm{r}^{\prime}(0)=\mathrm{r}_{1}(0)=\langle-1,1,2\rangle$
$(c)$
$\mathrm{r}^{\prime}=\mathrm{r}_{1} \times \mathrm{r}_{2}^{\prime}+\mathrm{r}_{1}^{\prime} \times \mathrm{r}_{2}$
$\langle-11,2\rangle \times\langle 4,0,2\rangle+\langle 1,2,1\rangle \times\{1,0,1\rangle=(0,10,-2)=\mathrm{r}^{\prime}(0)$
$(d)$
$r_{1}(t) \cdot r_{2}^{\prime}(t)+r_{1}^{\prime}(t) \cdot r_{2}(t)=f^{\prime}(t)$
$0+2=2=f^{\prime}(0)$
