# How do you use the product rule to differentiate y=(x+1)^2(2x-1)?

Mar 26, 2018

So I also need to use chain rule on ${\left(x + 1\right)}^{2}$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = u ' v + v ' u$
$u ' = 2 \left(x + 1\right) \cdot 1$
$v ' = 2$

$u = {\left(x + 1\right)}^{2}$
$v = \left(2 x - 1\right)$

subbing into the product rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(2 x + 1\right) \cdot \left(2 x - 1\right) + 2 {\left(x + 1\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(4 {x}^{2} - 1\right) + 2 \left({x}^{2} + 2 x + 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 {x}^{2} - 2 + 2 {x}^{2} + 4 x + 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 10 {x}^{2} + 4 x$

Mar 26, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {\left(x + 1\right)}^{2} + 2 \left(x + 1\right) \left(2 x - 1\right)$
or
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{3} + 8 {x}^{2} + 4 x - 2$

#### Explanation:

We know that a product is to things multiplied by each other so ${\left(x + 1\right)}^{2}$ and $\left(2 x - 1\right)$ are separate products

$u = {\left(x + 1\right)}^{2}$
$u ' = 2 \left(x + 1\right) \cdot 1$

$v = 2 x - 1$
$v ' = 2 x$

The product rule is $\frac{\mathrm{dy}}{\mathrm{dx}} = u v ' + v u '$

so it is

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {\left(x + 1\right)}^{2} + 2 \left(x + 1\right) \left(2 x - 1\right)$

simplified

dy/dx=2(x+1) ((x(x+1)+(2x-1))
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x + 2\right) \left({x}^{2} + x + 2 x - 1\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x + 2\right) \left({x}^{2} + 3 x - 1\right)$

Further simplification

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{3} + 6 {x}^{2} - 2 x + 2 {x}^{2} + 6 x - 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{3} + 8 {x}^{2} + 4 x - 2$