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

1 Answer
Sep 7, 2015

d/dx (x-1)(x-2)(x-3)= 3x^2 - 12x + 11

Explanation:

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.