# How do you differentiate  f(x)=(x-1)(x-2)(x-3) using the product rule?

Hey there!

Generally, to differentiate a product function, the formula is as follows:

$f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$

Where the derivative is:

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

#### Explanation:

In your example, this follows the general formula of product rule however, there is one extra term. To make up for this, we have to alter the general formula slightly. You now have to compensate for the third term.

So if:

$f \left(x\right) = g \left(x\right) \cdot h \left(x\right) \cdot i \left(x\right)$

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

In essence, you have to make sure you take the derivative of each term, then multiply by what you're not taking the derivative of!

$f \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)$

Following the general formula as "derived" above:

$f ' \left(x\right) = \left(1\right) \left(x - 2\right) \left(x - 3\right) + \left(1\right) \left(x - 1\right) \left(x - 3\right) + \left(1\right) \left(x - 2\right) \left(x - 1\right)$

Simplifying everything:

$f ' \left(x\right) = \left({x}^{2} - 5 x + 6\right) + \left({x}^{2} - 4 x + 3\right) + \left({x}^{2} - 3 x + 2\right)$

$f ' \left(x\right) = 3 {x}^{2} - 12 x + 11$

Therefore, the derivative (simplified) is: $f ' \left(x\right) = 3 {x}^{2} - 12 x + 11$

You could have stopped at the line before "Simplifying everything:", that is perfectly correct. If you can and it's not too difficult, it's probably better to simplify in case you need to do some direct substitution afterwards!

Hopefully this is clear and helpful! :)