Question

How do you find the derivative of `(4x+1)^2(1-x)^3`?

Answer 1

Use product and chain rule

Explanation:

`f'(x)=u'v+v'u`
Let `u=(4x+1)^2, v=(1-x)^3`
`u'=8(4x+1), v'=-3(1-x)^2`
`f'(x)=8(4x+1)(1-x)^3-3(1-x)^2(4x+1)^2`
`=(4x+1)(1-x)^2(8(1-x)-3(4x+1))`
`=(4x+1)(1-x)^2(5-20x)`

Answer 2

`f'(x)=-3(4x+1)^2(1-x)^2+8(4x+1)(1-x)^3`

Explanation:

You have to use a combination of the Product Rule and Chain Rule .

`f'(x)=uv'+u'v`

`u=(4x+1)^2`
`u'=2(4x+1)*4 =>` APPLY Chain Rule

`v=(1-x)^3`
`v'=3(1-x)^2*-1 =>` APPLY Chain Rule

`f'(x)=(4x+1)^2 3(1-x)^2*-1+2(4x+1)*4(1-x)^3`

`f'(x)=-3(4x+1)^2(1-x)^2+8(4x+1)(1-x)^3`