Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises - Page 901: 49

$f'(2)=35$

As we are given that $f(t)=u(t) \cdot v(t), u(2)=\lt 1,2,-1 \gt,u'(2)=\lt 3,0,4 \gt$ and $v(t)=\lt t, t^2,t^3 \gt$ Use 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 $v'(2)=\lt 1,4,12 \gt$ $f'(2)=\lt 3,0,4 \gt \cdot \lt 2,4,8 \gt+\lt 1,2,-1 \gt \cdot \lt 1,4,12\gt=6+0+32+1+8-12=35$