Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 27

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)