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
Consider, $u=\lt f(t), g(t), h(t) \gt$
a) Apply product rule to get $\dfrac{d(cu)}{dt}=\dfrac{d(c)}{dt}\lt f(t), g(t), h(t) \gt=c\dfrac{d)}{dt}\lt f(t), g(t), h(t) \gt$
Thus, $\dfrac{d(cu)}{dt}=c\dfrac{d(u)}{dt}$
b) Apply product rule to get $\dfrac{d(F(t)u)}{dt}=\dfrac{d}{dt}(f(t)\lt f(t), g(t), h(t) \gt)=F(t)\lt f'(t), g'(t), h'(t) \gt+F'(t)\lt f(t), g(t), h(t) \gt$
Thus, $\dfrac{d(F(t)u)}{dt}=F(t)u'+F'(t)u$
Hence, a) $\dfrac{d(cu)}{dt}=c\dfrac{d(u)}{dt}$
b) $\dfrac{d(F(t)u)}{dt}=F(t)u'+F'(t)u$
