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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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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