Compute the derivative of the vector-valued function r(t)=⟨sin⁡2t,et2⟩.

Step 1
In order to compute the derivative of the vector-valued function:
Differentiate the vector function on a component-by-component basis using the chain rule
Step 2
We differentiate the vector function on a component-by-component basis using the chain rule. This yields:
${\displaystyle {\left(\sin 2t\right)}^{{}^{\prime }}=\cos 2t\cdot {\left(2t\right)}^{{}^{\prime }}=2\cos 2t}$
${\displaystyle {\left(e^{t^2}\right)}^{{}^{\prime }}=e^{t^2}\cdot {\left(t^2\right)}^{{}^{\prime }}=2t{e}^{t^2}}$.
So the derivative is given by
${\displaystyle r^{{}^{\prime }}\left(t\right)=⟨2\cos 2t,2t{e}^{t^2}⟩.}$
Step 3
Hence, the final answer is:
${\displaystyle r^{{}^{\prime }}\left(t\right)=⟨2\cos 2t,2t{e}^{t^2}⟩.}$