# How do you differentiate (x^2-1)(2x)-(x^2-1)(2x)?

Aug 3, 2015

The answer is $\frac{d}{\mathrm{dx}} \left[\left({x}^{2} - 1\right) \left(2 x\right) - \left({x}^{2} - 1\right) \left(2 x\right)\right] = 0$

#### Explanation:

If $f \left(x\right) = \left({x}^{2} - 1\right) \left(2 x\right) - \left({x}^{2} - 1\right) \left(2 x\right)$ and $x$ exists, then
$f \left(x\right) = 0$.

Luckily the derivative of a constant is zero, so

$\frac{d}{\mathrm{dx}} \left[\left({x}^{2} - 1\right) \left(2 x\right) - \left({x}^{2} - 1\right) \left(2 x\right)\right] = \frac{d}{\mathrm{dx}} \left[0\right] = 0.$

You could also practice your rules of derivatives if you like, so that

$\frac{d}{\mathrm{dx}} \left[\left({x}^{2} - 1\right) \left(2 x\right) - \left({x}^{2} - 1\right) \left(2 x\right)\right]$

By the Product Rule
$= \frac{d}{\mathrm{dx}} \left[\left({x}^{2} - 1\right) \left(2 x\right)\right] - \frac{d}{\mathrm{dx}} \left[\left({x}^{2} - 1\right) \left(2 x\right)\right]$
$= \left[\left({x}^{2} - 1\right) 2 + 2 x \left(2 x\right)\right] - \left[\left({x}^{2} - 1\right) 2 + 2 x \left(2 x\right)\right]$
$= \left[2 {x}^{2} - 2 + 4 {x}^{2}\right] - \left[2 {x}^{2} - 2 + 4 {x}^{2}\right]$
$= \left[6 {x}^{2} - 2\right] - \left[6 {x}^{2} - 2\right]$
$= \left(6 {x}^{2} - 6 {x}^{2}\right) + \left(2 - 2\right)$
$= 0 + 0 = 0$