To differentiate the function f(x) = (3-1/x)/(x-1)/

We multiply both numerator and denominator on right side by x and then we get:

f(x) = (3x-1)/(x*2-x)

f(x) is in u(x)/v(x) form.

(u(x)/v(x))' = {u'(x)v(x) - u(x)v'(x)}/(v(x))^2.

Therefore f'(x) = {(3x-1)'(x^2-x) - (3x-1)(x^2-x)'}/((x^2-x)^2

f'(x) = {3(x^2-x)- (3x-1)(2x-1)}/{x^2-x}^2.

f'(x) = {3x^2 -3x - 6x^2 +5x-1}/(x^2-x)^2.

f'(x) = - {3x^2 -2x+1}/(x^2-x)^2.

f'(x) = - (3x^2-2x+1)/{x^2(x-1)^2}.

We'll differentiate with respect to x:

df/dx = d/dx {[3-(1/x)] / (x-1)}

df/dx = d/dx [3/(x-1)] - d/dx [1/x(x-1)]

d/dx [3/(x-1)] = [(x-1)*d/dx(3) - 3*d/dx(x-1)]/(x-1)^2

d/dx [3/(x-1)] = [0*(x-1) - 3*1]/(x-1)^2

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

We'll differentiate the term d/dx [1/x(x-1)]:

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

d/dx [1/(x^2 - x)] = [(x^2 - x)*d/dx(1) - 1*d/dx(x^2 - x)]/x^2*(x-1)^2

d/dx [1/(x^2 - x)] = -(2x-1)/x^2*(x-1)^2 (2)

df/dx = (1) - (2)

df/dx = - 3/(x-1)^2 + (2x-1)/x^2*(x-1)^2

**df/dx = (2x - 1 - 3x^2)/x^2*(x-1)^2**