Answer
$\langle 4e^{4t+9},8e^{8t+18},0\rangle $
Work Step by Step
Since $ r(t)=\langle e^{t}, e^{2t},4\rangle, \quad g(t)=4t+9 $, then we have $$\frac{d}{dt}(r(g(t)))=r'(g(t)) g'(t)=4\langle e^{4t+9}, 2e^{8t+18},0\rangle=\langle 4e^{4t+9},8e^{8t+18},0\rangle.$$
