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

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Answer 1 · 2018-03-26T11:19:23

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

$dy/dx = u'v + v'u$
$u' = 2(x+1) * 1$
$v' = 2$

$u=(x+1)^2$
$v=(2x-1)$

subbing into the product rule.

$dy/dx = 2(2x+1) * (2x-1) + 2(x+1)^2$

$dy/dx = 2(4x^2-1) + 2(x^2+2x+1)$

$dy/dx = 8x^2-2 + 2x^2+4x+2$

$dy/dx = 10x^2+4x$

Answer 2 · 2018-03-26T11:26:31

$dy/dx=2x(x+1)^2+2(x+1)(2x-1)$
or
$dy/dx=2x^3+8x^2+4x-2$

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

$u=(x+1)^2$
$u'=2(x+1)*1$

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

The product rule is $dy/dx=uv'+vu'$

so it is

$dy/dx=2x(x+1)^2+2(x+1)(2x-1)$

simplified

$dy/dx=2(x+1) ((x(x+1)+(2x-1))$
$dy/dx=(2x+2) (x^2+x+2x-1)$
$dy/dx=(2x+2) (x^2+3x-1)$

Further simplification

$dy/dx=2x^3+6x^2-2x+2x^2+6x-2$
$dy/dx=2x^3+8x^2+4x-2$

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