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

We know that if f= u/v , then f'= (u'v-uv')/v^2

Now we will assume that:

u= x^2 ==> u'=2x

v= x-2 ==> v' =1

Then f'(x) = [2x(x-2)- x^2(1)]/(x-2)^2

= (2x^2 -4x -x^2 )/(x-2)^2

= (x^2 -4x)/(x-2)^2

Now we will substitute with x=1

f'(1) = (1-4)/(1-2)^2

= -3/ 1= -3

Then f'(1)= -3

The function `f(x)= x^2 / (x-2)` .

Use the property `x^-1 = 1/x` . This allows us to write the given function `f(x)= x^2 / (x-2)` as `f(x)= x^2*(x-2)^-1` .

If `y = f(x)*g(x)` , `y' = f'(x)*g(x) + f(x)*g'(x)` and if `y = x^n, y' = n*x^(n-1)`

For the given function `f(x)= x^2*(x-2)^-1`

`f'(x) = (x^2)'*(x - 2)^-1 + x^2*((x-2)^-1)'`

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

At x = 1, `2x*(x - 2)^-1 + x^2*-1*(x - 2)^-2` = `2*1*(1 - 2)^-1 + 1^2*-1*(1 - 2)^-2`

= `2*-1 + -1*(1/1)`

= -2 - 1

= -3

The derivative of `f(x)= x^2 / (x-2)` at x = 1 is -3

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

D= R-{2} ;than we could write

f(x)'= (x^2)'(x-2)-[(x-2)'(x^2)]/(x-2)^2

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

f(x)'=2x^2-4x-x^2/(x-2)^2

f(x)'=x^2-4x/(x-2)^2

f(x)'=x(x-4)/(x-2)^2

finally f(1)'=1(1-4)/1-2)^2

Another method to calculate a derivative of a function in a point is:

f'(1) = limit [f(x)-f(1)]/(x-1) = lim [x^2/(x-2) + 1]/(x-1)

f'(1) = lim (x^2 + x - 2)/(x-2)(x-1) = 0/0

If we'll substitute x by 1 into the numerator expression, we'll cancel it, so x=1 is the root of both, numerator and denominator. We'll write the numerator as a product.

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

f'(1) = lim (x-1)(x+2)/(x-1)(x-2) = lim (x+2)/(x-2)

**f'(1) = (1+2)/(1-2) = 3/-1 = -3**

f(x) = x^2/(x-2). To find f'(10.

Solution:

We first find f'(x) and then find the value of f'(x) at x=1.

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

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

= (x(x-2) + 2(x-2)+4)/(x-2)

=x +2 +4/(x-2)

f'(x) = {x +2 +4(x-2)}'.

= (x)'+(2)' +(4/(x-2))'. Using (d/dx)x^n = nx^(n-1) ..

= 1 +0 -2/(x-2)^2 , as (d/dx) (ax+b)^n = n(ax+b)^(n-1)* a. Therefore,

f'(x) = 1 -2/(x-2)^2.

f'(1) = 1 -2/(1-2)^2 = 1 -2/1 = -1