Verified answer

f'(x) = [(x²)'(x-1) - x²(x-1)']/(x-1)² = [2x(x-1) - x²(1)]/(x-1)² = (2x² - x² - 2x)/(x-1)² = x(x-2)/(x-1)²

...

By the quotient rule, f '(x)

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

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

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

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