Answer
$\dfrac{d}{dt}[u(f(t))]=f'(t)u'(f(t))$
Work Step by Step
Since, $\dfrac{d}{dt}[u(f(t))]=\dfrac{d}{dt}[u_1'(f(t))i+u_2'(f(t))j+u_3'(f(t))k]$
This gives: $=u_1'(f(t))f'(t)i+u_2'(f(t))f'(t)j+u_3'(f(t))f'(t)k$
or, $=f'(t)[u_1'(f(t))i+u_2'(f(t))j+u_3'(f(t))k]$
or, $=f'(t)u'(f(t))$
Hence, the result.
$\dfrac{d}{dt}[u(f(t))]=f'(t)u'(f(t))$
