What is f'(0) if f(x)=1/(x-1)(x-2)?

Given that:

f(x) =
1/(x-1)(x-2)

We need to find
f'(0)

First we need to find the first derivative
f'(x).

Let f(x) = 1/(x-1) *(1/(x-2) =
u*v

such that:

u = 1/(x-1) 
==> u' = -1/(x-1)^2

v= 1/(x-2) ==> v' =
-1/(x-2)^2

==> We will use the product rule to
determine f'(x).

==> f'(x) = u'*v +
u*v'

              = (-1/(x-1)^2 * 1/(x-2) + (1/(x-1)
*(-1/(x-2)^2

              = -1/(x-2)(x-1)^2 -
1/(x-1)(x-2)^2

               = ( -(x-2) - (x-1) /[
(x-1)(x-2)]^2

                =
(-x+2-x+1)/[(x-1)(x-2)]^2

               =
(-2x+3)/[(x-1)(x-2)]^2

==> f'(x) =
(-2x+3)/[(x-1)(x-2)]^2

==> f'(0) =  3/
(-1)(-2)]^2

                =
3/4

==> f'(0) =
3/4