Answer
$\dfrac{d}{dt}[u(t) \cdot v(t))]=2t \cos t+2 \sin t -2 \cos t \sin t$
Work Step by Step
As we are given that $u(t)=\lt \sin t,\cos t, t \gt$ and $v(t)=\lt t, \cos t, \sin t \gt$
Need to prove $\dfrac{d}{dt}[u(t) \cdot v(t))]$.
$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 $
Use product formula: $\dfrac{d}{dt}[u(t) \cdot v(t)]=u'(t) \cdot v(t)+u(t) \cdot v'(t)$
We have $\dfrac{d}{dt}[u(t) \cdot v(t)]=\lt \cos t,-\sin t, 1 \gt \cdot \lt t, \cos t, \sin t \gt+\lt \sin t,\cos t, t \gt \cdot \lt 1, -\sin t, \cos t \gt$
or, $=2t \cos t+2 \sin t -2 \cos t \sin t$
Hence, the result.
$\dfrac{d}{dt}[u(t) \cdot v(t))]=2t \cos t+2 \sin t -2 \cos t \sin t$
