Answer
$$\frac{d}{dt}(t^4r_1(t))
=\langle 6 t^5,7t^6,5t^4 \rangle .$$
Work Step by Step
Since $$ r_1(t)=\langle t^2,t^3,t \rangle $$
then, we have
$$\frac{d}{dt}(t^4r_1(t))=\frac{d}{dt}(t^4)r_1(t)) +t^4 \frac{d}{dt}( r_1(t))\\
=4t^3\langle t^2,t^3,t \rangle+t^4 \langle 2t,3t^2,1 \rangle\\
=\langle 6 t^5,7t^6,5t^4 \rangle .$$
