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

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

=> [x^2 - 3x + 2]^-1

Using the chain rule

f'(x) = -[x^2 - 3x + 2]^-2 * ( 2x - 3)

=> -(2x - 3)/ (x^2-3x + 2)^2

f'(0) = -(0 - 3) / (0 + 2)^2

=> 3 / 4

**Therefore f'(0) = 3/4**

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**

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