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

`f'(3)=100`

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`