Answer
$\dfrac{d}{dt}[u(t) \times v(t))]=(\cos 2t+t \sin t-\cos t)i+(2t-\sin 2t)j+(\cos 2t+t\sin t-\cos t)k$
Work Step by Step
$u(t)=\lt \sin t,\cos t, t \gt$ and $v(t)=\lt t, \cos t, \sin t \gt$
Our aim is to prove: $\dfrac{d}{dt}[u(t) \times v(t))]$.
$u(t)=\lt \sin t,\cos t, t \gt \implies u'(t)=\lt \cos t,-\sin t, 1 \gt $
and
$v(t)=\lt t, \cos t, \sin t \gt \implies v'(t)=\lt 1, -\sin t, \cos t \gt $
Since, we have $\dfrac{d}{dt}[u(t) \times v(t)]=u'(t) \times v(t)+u(t) \times v'(t)$
Thus, $\dfrac{d}{dt}[u(t) \times v(t)]=\begin{vmatrix}i&j&k \\cos t&-\sin t&1 \\t&\cos t&\sin t\end{vmatrix}+\begin{vmatrix}i&j&k \\\sin t&\cos t&t \\1&-\sin t&\cos t\end{vmatrix}$
$=(-sin^2t-cost)i+(t-sintcost)j+(cos^2t+tsint)k+(cos^2t+tsint)i+(t-sintcost)j+(-sin^2t-tcost)k$
$=(cos^2t-sin^2t+tsint-cost)i+2(t-sintcost)j+(cos^2t-sin^2t+tsint-cost)k$
$=(cos2t+tsint-cost)i+(2t-sin2t)j+(cos2t+tsint-cost)k$
Hence, $\dfrac{d}{dt}[u(t) \times v(t))]=(\cos 2t+t \sin t-\cos t)i+(2t-\sin 2t)j+(\cos 2t+t\sin t-\cos t)k$
