1. **State the problem:** Find the derivative of the function $$f(x) = 1 - \frac{\ln((x-1)^2)}{x-1}$$. 2. **Recall the formulas:** - Derivative of $$\ln(u)$$ is $$\frac{u'}{u}$$. - Quotient rule: If $$f(x) = \frac{g(x)}{h(x)}$$, then $$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$$. 3. **Simplify the function inside the logarithm:** $$\ln((x-1)^2) = 2\ln|x-1|$$. 4. **Rewrite the function:** $$f(x) = 1 - \frac{2\ln|x-1|}{x-1}$$. 5. **Set:** $$g(x) = 2\ln|x-1|$$ and $$h(x) = x-1$$. 6. **Find derivatives:** - $$g'(x) = 2 \cdot \frac{1}{x-1} = \frac{2}{x-1}$$. - $$h'(x) = 1$$. 7. **Apply quotient rule:** $$\left(\frac{g}{h}\right)' = \frac{g'h - gh'}{h^2} = \frac{\frac{2}{x-1}(x-1) - 2\ln|x-1| \cdot 1}{(x-1)^2} = \frac{2 - 2\ln|x-1|}{(x-1)^2}$$. 8. **Derivative of the whole function:** $$f'(x) = 0 - \frac{2 - 2\ln|x-1|}{(x-1)^2} = -\frac{2 - 2\ln|x-1|}{(x-1)^2} = \frac{2\ln|x-1| - 2}{(x-1)^2}$$. 9. **Final answer:** $$f'(x) = \frac{2\ln|x-1| - 2}{(x-1)^2}$$.