Answer
$\dfrac{d}{dt}[f(t)u(t)]=f'(t)u(t)+f(t)u'(t)$
Work Step by Step
Our aim is to prove: $\dfrac{d}{dt}[f(t)u(t)]=f'(t)u(t)+f(t)u'(t)$ ...(1)
Suppose $u(t)=u_1(t)i+u_2(t)j+u_3(t)k$ and
Take the left side of the equation (1).
$\dfrac{d}{dt}[f(t)u(t)]=\dfrac{d}{dt}[(f(t)u_1(t)i))(f(t)u_2(t)j)+(f(t)u_3(t)k)]$
$=f'(t)[u_1(t)i+u_2(t)j+u_3(t)k]+f(t)[u_1'(t)i+u_2'(t)j+u_3'(t)k]$
$=f'(t)u(t)+f(t)u'(t)$
Hence, $\dfrac{d}{dt}[f(t)u(t)]=f'(t)u(t)+f(t)u'(t)$
