Answer
a) $\dfrac{d(cu)}{dt}=c\dfrac{d(u)}{dt}$
b) $\dfrac{d(F(t)u)}{dt}=F(t)u'+F'(t)u$
Work Step by Step
a. Let us consider, $u=\lt f(t), g(t), h(t) \gt$
Need to use product rule such as:
$\dfrac{d(cu)}{dt}=\dfrac{d(c)}{dt}\lt f(t), g(t), h(t) \gt=c\dfrac{d}{dt}[\lt f(t), g(t), h(t) \gt]$
Therefore, $\dfrac{d(cu)}{dt}=c\dfrac{d(u)}{dt}$
b.
Let us consider, $u=\lt f(t), g(t), h(t) \gt$
Need to use product rule such as:$\dfrac{d(F(t)u)}{dt}=\dfrac{d}{dt} [f(t)\lt f(t), g(t), h(t) \gt ]=F(t)\lt f'(t), g'(t), h'(t) \gt+F'(t)\lt f(t), g(t), h(t) \gt$
Therefore, $\dfrac{d[F(t)(u)]}{dt}=F(t)u'+F'(t)u$
