## Calculus 8th Edition

$\dfrac{d}{dt}[f(t)u(t)]=f'(t)u(t)+f(t)u'(t)$
Let us consider $u(t)=u_1(t)i+u_2(t)j+u_3(t)k$ $\dfrac{d}{dt}[f(t)u(t)]=\dfrac{d}{dt}[(f(t)u_1(t)i))(f(t)u_2(t)j)+(f(t)u_3(t)k)]$ or, $=f'(t)[u_1(t)i+u_2(t)j+u_3(t)k]+f(t)[u_1'(t)i+u_2'(t)j+u_3'(t)k]$ or, $=f'(t)u(t)+f(t)u'(t)$ Hence, the result. $\dfrac{d}{dt}[f(t)u(t)]=f'(t)u(t)+f(t)u'(t)$