How to differentiate $\frac{x}{1-\ln(x-1)}$? I'm working on the following problem found in James Stewart's Calculus Early Transcendentals, 7th Ed., Page 223, Exercise 27. I'd just like to know where my work had gone wrong?
Please differentiate:  $f(x)=\frac{x}{1-\ln(x-1)}$
My work is below. First I apply quotient rule and chain rules.
$$f'(x)=\frac{\left(1-\ln(x-1)\right)(1)-\left((x)(-\left(\frac{1}{x-1}\right)(1)\right)}{\left(1-\ln(x-1)\right)^2}$$
My algebraic simplification:
$$f'(x)=\frac{\left(1-\ln(x-1)\right)-(x-\left(\frac{1}{x-1}\right))}{\left(1-\ln(x-1)\right)^2}$$
$$f'(x)=\frac{\left(1-\ln(x-1)\right)+(-x+\left(\frac{1}{x-1}\right))}{\left(1-\ln(x-1)\right)^2}$$
$$f'(x)=\frac{\left(1-\ln(x-1)\right)+(-x+\left(\frac{1}{x-1}\right))}{\left(1-\ln(x-1)\right)^2}$$
$$f'(x)=\frac{1-\ln(x-1)-x+1}{\left(1-\ln(x-1)\right)^2(x-1)}$$
$$f'(x)=\frac{-x+2-\ln(x-1)}{\left(1-\ln(x-1)\right)^2(x-1)}$$
However the solution is:
$$f'(x)=\frac{\left(2x-1-(x-1)\ln(x-1)\right)}{(1-\ln(x-1))^2(x-1)}$$
Just need to know where my work is incorrect. Thank you for your help!

The negative sign you have in the second term in the numerator (by applying the quotient rule) is a multiplication. You carried that out as a difference, which is wrong.
