## Calculus 8th Edition

$\dfrac{d}{dt}[u(t) \cdot v(t))]=2t \cos t+2 \sin t -2 \cos t \sin t$
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. $\dfrac{d}{dt}[u(t) \cdot v(t))]=2t \cos t+2 \sin t -2 \cos t \sin t$