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

Oct 5, 2016

Use product and chain rule

#### Explanation:

$f ' \left(x\right) = u ' v + v ' u$
Let $u = {\left(4 x + 1\right)}^{2} , v = {\left(1 - x\right)}^{3}$
$u ' = 8 \left(4 x + 1\right) , v ' = - 3 {\left(1 - x\right)}^{2}$
$f ' \left(x\right) = 8 \left(4 x + 1\right) {\left(1 - x\right)}^{3} - 3 {\left(1 - x\right)}^{2} {\left(4 x + 1\right)}^{2}$
$= \left(4 x + 1\right) {\left(1 - x\right)}^{2} \left(8 \left(1 - x\right) - 3 \left(4 x + 1\right)\right)$
$= \left(4 x + 1\right) {\left(1 - x\right)}^{2} \left(5 - 20 x\right)$

Oct 5, 2016

$f ' \left(x\right) = - 3 {\left(4 x + 1\right)}^{2} {\left(1 - x\right)}^{2} + 8 \left(4 x + 1\right) {\left(1 - x\right)}^{3}$

#### Explanation:

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

$f ' \left(x\right) = u v ' + u ' v$

$u = {\left(4 x + 1\right)}^{2}$
$u ' = 2 \left(4 x + 1\right) \cdot 4 \implies$ APPLY Chain Rule

$v = {\left(1 - x\right)}^{3}$
$v ' = 3 {\left(1 - x\right)}^{2} \cdot - 1 \implies$ APPLY Chain Rule

$f ' \left(x\right) = {\left(4 x + 1\right)}^{2} 3 {\left(1 - x\right)}^{2} \cdot - 1 + 2 \left(4 x + 1\right) \cdot 4 {\left(1 - x\right)}^{3}$

$f ' \left(x\right) = - 3 {\left(4 x + 1\right)}^{2} {\left(1 - x\right)}^{2} + 8 \left(4 x + 1\right) {\left(1 - x\right)}^{3}$