How do you differentiate y=(x+1)^2(2x-1)?

Aug 13, 2017

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

Explanation:

$y = {\left(x + 1\right)}^{2} \left(2 x + 1\right)$

Use the product rule and chain rule to differentiate $y$

Product rule:

$\frac{d}{\mathrm{dx}} \left(p q\right) = q p ' + p q '$

Chain rule:

$\frac{d}{\mathrm{dx}} \left({\left[f \left(x\right)\right]}^{n}\right) = n {\left[f \left(x\right)\right]}^{n - 1} f ' \left(x\right)$

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

Aug 13, 2017

$6 x \left(x + 1\right)$

Explanation:

Actually, there are mainly 2 methods of solving your question,

1. Either simplifying the equation and differentiating it

2. Or differentiating first and then simplifying it.

Obviously, the first one is the easiest for beginners,

$\Rightarrow y = {\left(x + 1\right)}^{2} \left(2 x - 1\right)$

Simplifying this, I get,
$\Rightarrow y = 2 {x}^{3} + 3 {x}^{2} - 1$

Now,
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{2} + 6 x$

$\therefore , \frac{\mathrm{dy}}{\mathrm{dx}} = 6 x \left(x + 1\right)$

ENJOY MATHS !!!!!!!!!