This is a plot of `f(x) `, we will find the domain of `f(x) `, notice the missing values represented by circles in the plot.

Given `f(x) = x/(1-ln(x-1))` , we know that the denominator `1-ln(x-1)` cannot be zero. This means that ln(x-1)`!=1`. But when is ln(x-1)`=1`, that happens to be when `x-1=e=>x=1+e~~3.7 ` on the plot. Now `ln(x-1) ` can only take values of `x>1 ` since `x-1>0 ` for the `ln ` function to work. Thus, the domain of `f(x) ` is `x in (1,1+e) uu (1+e,+oo]`.

Turning over to the derivative of `f(x) `. Use the Quotient rule. To remind you it's `(f'*g-f*g')/(g^2) ` . Thus, `f'(x)=(1*(1-ln(x-1))-x*(-(1)/(x-1)))/((1-ln(x-1))^2) ` . Here we have used the fact that `(ln(x-1))'=(1)/(x-1) ` .

Going a bit further with simplification, `f'(x)=(1)/(1-ln(x-1))+(x)/((x-1)*(1-ln(x-1))^2) `