1. **State the problem:** Find the derivative of the function $$f(x) = \frac{\ln(x)}{(x-1)^2}$$. 2. **Recall the formula:** To differentiate a quotient $$\frac{u}{v}$$, use the quotient rule: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $$u = \ln(x)$$ and $$v = (x-1)^2$$. 3. **Find derivatives of numerator and denominator:** - $$u' = \frac{1}{x}$$ (derivative of $$\ln(x)$$) - $$v' = 2(x-1)$$ (derivative of $$(x-1)^2$$) 4. **Apply the quotient rule:** $$f'(x) = \frac{\frac{1}{x} \cdot (x-1)^2 - \ln(x) \cdot 2(x-1)}{(x-1)^4}$$ 5. **Simplify the numerator:** $$\frac{1}{x}(x-1)^2 - 2\ln(x)(x-1) = \frac{(x-1)^2}{x} - 2(x-1)\ln(x)$$ 6. **Write the final derivative:** $$f'(x) = \frac{\frac{(x-1)^2}{x} - 2(x-1)\ln(x)}{(x-1)^4} = \frac{(x-1)^2 - 2x(x-1)\ln(x)}{x(x-1)^4}$$ 7. **Further simplification:** $$f'(x) = \frac{(x-1)^2 - 2x(x-1)\ln(x)}{x(x-1)^4} = \frac{(x-1) - 2x\ln(x)}{x(x-1)^3}$$ **Answer:** $$\boxed{f'(x) = \frac{(x-1) - 2x\ln(x)}{x(x-1)^3}}$$