Answer
$f'(2)=35$
Work Step by Step
Given: $f(t)=u(t) \cdot v(t), u(2)=\lt 1,2,-1 \gt,u'(2)=\lt 3,0,4 \gt$ and $v(t)=\lt t, t^2,t^3 \gt$
Our aim is to calculate $f'(2)$.
In order to find this, we will use the dot product rule.
$f'(t)=u'(t) \cdot v(t)+u(t) \cdot v'(t)$
$v(t)=\lt t, t^2,t^3 \gt \implies v'(t)=\lt 1, 2t,3t^2 \gt$
and $v'(2)=\lt 1,4,12 \gt$
Thus, $f'(2)=\lt 3,0,4 \gt \cdot \lt 2,4,8 \gt+\lt 1,2,-1 \gt \cdot \lt 1,4,12\gt$
$=6+0+32+1+8-12$
$=35$
Hence, $f'(2)=35$
