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