Answer
$$
R_v= -\frac{2v}{k}e^{-v^2/k}, \quad R_k= \frac{v^2}{k^2}e^{-v^2/k}.
$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(x^n)'=nx^{n-1}$
Since $ R=e^{-v^2/k}$, then by using the chain rule, we have
$$
R_v= -\frac{2v}{k}e^{-v^2/k}, \quad R_k= \frac{v^2}{k^2}e^{-v^2/k}.
$$
