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

Jun 5, 2016

The rule of product tells you that the derivative of a product of two functions is

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right)\right] = \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} g \left(x\right) + f \left(x\right) \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}$.

In our case the the product is triple, so we will first consider
$\left(x - 1\right) \cdot \left[\left(x - 2\right) \left(x - 3\right)\right]$ as a product, then we will do the second part repeating the rule again.

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\left(x - 1\right) \left[\left(x - 2\right) \left(x - 3\right)\right]\right)$

$= \frac{d \left(x - 1\right)}{\mathrm{dx}} \left[\left(x - 2\right) \left(x - 3\right)\right] + \left(x - 1\right) \frac{d \left[\left(x - 2\right) \left(x - 3\right)\right]}{\mathrm{dx}}$

$= \left(x - 2\right) \left(x - 3\right) + \left(x - 1\right) \frac{d \left[\left(x - 2\right) \left(x - 3\right)\right]}{\mathrm{dx}}$

now we reapply the same rule to the next product

$= \left(x - 2\right) \left(x - 3\right) + \left(x - 1\right) \left(\frac{d \left(x - 2\right)}{\mathrm{dx}} \left(x - 3\right) + \left(x - 2\right) \frac{d \left(x - 3\right)}{\mathrm{dx}}\right)$

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

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