Question

Findf '(3), where f(t) = u(t) · v(t), u(3) =1, 2, −2, u'(3) =8, 1, 4,andv(t) =t, t2, t3.

Answer 1

Answer:

$f'(3)=100$

Step-by-step explanation:

Given:

$f(t)=u(t)\cdot v(t)\\u(3)=\left ( 1,2,-2 \right )\\u'\left ( 3 \right )=\left ( 8,1,4 \right )\\v(t)=\left ( t,t^{2},t^{3} \right )$

To find: $f'(3)$

Solution:

$v(t)=\left ( t,t^{2},t^{3} \right )$

At $t=3;$

$v(3)=(3,3^{2},3^{3} )=(3,9,27)$

Differentiate with respect to t

$v'(t)=\left ( 1,2t,3t^{2} \right )$

At $t=3;$

$v'(3)=\left ( 1,2(3),3(3)^{2} \right )=\left ( 1,6,27 \right )$

Using product rule, differentiate $f(t)=u(t)\cdot v(t)$ with respect to $t$

$f'(t)=u'(t)\cdot v(t)+u(t)\cdot v'(t)$

At $t=3;$

$f'(3)=u'(3)\cdot v(3)+u(3)\cdot v'(3)\\=\left ( 8,1,4 \right )\cdot \left ( 3,9,27 \right )+\left ( 1,2,-2 \right )\cdot \left ( 1,6,27 \right )\\=24+9+108+1+12-54\\=100$