`f(x) = log_2 (x^2/(x-1))`
The derivative formula of a logarithm is
`d/dx [ log_a (u)] = 1/(ln(a) * u) * (du)/dx`
Applying this formula, the derivative of the function will be:
`f'(x) = d/dx [ log_2 (x^2/(x-1))]`
`f'(x) = 1/(ln(2) * (x^2/(x-1))) * d/dx(x^2/(x-1))`
`f'(x) = (x-1)/(x^2ln(2)) * d/dx(x^2/(x-1))`
To take the derivative of `x^2/(x-1)` , apply the quotient rule `d/dx[(f(x))/(g(x))] = (g(x)*f'(x) - f(x)*g'(x))/[g(x)]^2` .
`f'(x) = (x-1)/(x^2ln(2)) * ((x-1)*2x - x^2*1)/(x-1)^2`
`f'(x) = (x-1)/(x^2ln(2)) * (2x^2-2x - x^2)/(x-1)^2`
`f'(x) = (x-1)/(x^2ln(2)) * (x^2 - 2x)/(x-1)^2`
`f'(x) = (x-1)/(x^2ln(2)) * (x(x-2))/(x-1)^2`
`f'(x) = 1/(xln(2))*(x-2)/(x-1)`
`f'(x)=(x-2)/(x(x-1)ln(2))`
Therefore, the derivative of the function is `f'(x)=(x-2)/(x(x-1)ln(2))` .
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