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 following is correct:
$$f'(x)=\frac{\left(1-\ln(x-1)\right)(1)-\left((x)(-\left(\frac{1}{x-1}\right)(1)\right)}{\left(1-\ln(x-1)\right)^2}$$
The following is incorrect:
$$f'(x)=\frac{\left(1-\ln(x-1)\right)-(x-\left(\frac{1}{x-1}\right))}{\left(1-\ln(x-1)\right)^2}$$
On the right term in the numerator, you added $x$ and $-\frac{1}{x-1}$, but you were actually supposed to multiply them.
I'll go off of the top to finish. Simplify the numerator:
$$f'(x)=\frac{1-\ln(x-1)-\frac{-x}{x-1}}{\left(1-\ln(x-1)\right)^2}$$
Get rid of the double negative:
$$f'(x)=\frac{1-\ln(x-1)+\frac{x}{x-1}}{\left(1-\ln(x-1)\right)^2}$$
Multiply both the numerator and denominator by $x-1$. When you did this above, you did not multiply the $1$ and $-\ln(x-1)$ terms by $x-1$, which was another mistake, so remember to distribute the $x-1$ all the way through:
$$f'(x)=\frac{1(x-1)-\ln(x-1)(x-1)+x}{\left(1-\ln(x-1)\right)^2(x-1)}$$
Simplify the numerator:
$$f'(x)=\frac{2x-1-\ln(x-1)(x-1)}{\left(1-\ln(x-1)\right)^2(x-1)}$$
