# How do you differentiate f(x) = (x-1)(x-2) using the product rule?

Jun 8, 2016

The product rule states that for a function $f \left(x\right) = g \left(x\right) h \left(x\right)$, $f ' \left(x\right) = \left(g ' \left(x\right) \times h \left(x\right)\right) + \left(h ' \left(x\right) \times g \left(x\right)\right)$

#### Explanation:

In our case, let $f \left(x\right) = g \left(x\right) \times h \left(x\right)$

Therefore, $g \left(x\right) = x - 1$ and $h \left(x\right) = x - 2$

Let's differentiate both these functions.

By the power rule:

$g ' \left(x\right) = 1$

and

$h ' \left(x\right) = 1$

Now, we can apply the product rule.

$f ' \left(x\right) = \left(g ' \left(x\right) \times h \left(x\right)\right) + \left(h ' \left(x\right) \times g \left(x\right)\right)$

$f ' \left(x\right) = \left(\left(x - 2\right) \times 1\right) + \left(\left(x - 1\right) \times 1\right)$

$f ' \left(x\right) = x - 2 + x - 1$

$f ' \left(x\right) = 2 x - 3$

Therefore $\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 3$

Hopefully this helps!