Question

F(x) = (x ^ 2 + 2x + 7) ^ 3 then

Answer 1

`f^(\prime)(x)=6(x+1)(x^(2)+2x+7)^(2)`
`f^(\prime)(1)=1200`

Step-by-step explanation

Given

`f\mleft(x\mright)=(x^2+2x+7)^3`

To find the derivative, we have to apply the chain rule:

`[u(x)^n]^(\prime)=n\cdot u(x)^(n-1)\cdot u^(\prime)(x)`

Considering that in our case,

`u(x)=x^2+2x+7`
`u^(\prime)(x)=2x+2+0`

and n = 3, then:

`=3\cdot(x^2+2x+7)^(3-1)\cdot(2x+2)`

Simplifying:

`f^(\prime)(x)=3\cdot2(x+1)(x^2+2x+7)^2`
`f^(\prime)(x)=6(x+1)(x^2+2x+7)^2`

Finally, we have to replace 1 in each x in f'(x) to find f'(1):

`f^(\prime)(1)=6((1)+1)((1)^2+2(1)+7)^2`
`f^(\prime)(1)=6(1+1)(1+2+7)^2`
`f^(\prime)(1)=6(2)(10)^2`
`f^(\prime)(1)=6(2)(100)`
`f^(\prime)(1)=12(100)`
`f^(\prime)(1)=1200`