How do you differentiate `f(x)=(x-1)(x-2)(x-3)`?

One method is to expand the entire expression, and then differentiate using the power rule .

`d/dx (x-1)(x-2)(x-3)`
`= d/dx[(x^2 - 3x + 2) (x-3) ]`
`= d/dx(x^3 - 3x^2 + 2x - 3x^2 +9x-6 )`
`= d/dx(x^3 - 6x^2 + 11x - 6 )`
`= d/dx(x^3) - d/dx(6x^2) + d/dx(11x) - d/dx(6)`
`= 3x^2 - 12x + 11 - 0`
`= 3x^2 - 12x + 11`

Another method is to use the product rule, which I'll show here:
`d/dx f(x)g(x) = f(x) d/dx g(x) + g(x) d/dx f(x)`

We want to find `d/dx (x-1)(x-2)(x-3)`
We can apply the product rule in stages:

First, let's let our `f(x) = x-1` and `g(x) = (x-2)(x-3)`.

`d/dx (x-1)(x-2)(x-3)`
`= (x-1)d/dx[(x-2)(x-3)] + (x-2)(x-3)d/dx[x-1]`

The right side is easy enough: `d/dx[x-1] = (1-0) = 1`

So now we have:
`(x-1)d/dx[(x-2)(x-3)] + (x-2)(x-3)`

Now we apply the product rule again, this time with `f(x) = x-2` and `g(x) = x-3`.

`(x-1)d/dx[(x-2)(x-3)] + (x-2)(x-3)`

`= (x-1)[(x-2)d/dx(x-3) + (x-3)d/dx(x-2)] + (x-2)(x-3)`

`= (x-1)[(x-2) + (x-3)] + (x-2)(x-3)`

`= (x-1)(2x-5) + (x-2)(x-3)`

`= (2x^2 - 7x +5) + (x^2 - 5x + 6)`

`= 3x^2 - 12x + 11`

which is the same as our previous answer.