1. **State the problem:** Find the derivative of the function $$f(x) = \frac{\ln(x) - 1}{x - e}$$ 2. **Recall the quotient rule:** For a function $f(x) = \frac{g(x)}{h(x)}$, the derivative is $$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$$ 3. **Identify components:** - Numerator: $g(x) = \ln(x) - 1$ - Denominator: $h(x) = x - e$ 4. **Compute derivatives:** - $g'(x) = \frac{1}{x}$ because derivative of $\ln(x)$ is $\frac{1}{x}$ and derivative of constant $-1$ is 0. - $h'(x) = 1$ because derivative of $x$ is 1 and $e$ is constant. 5. **Apply quotient rule:** $$f'(x) = \frac{\frac{1}{x}(x - e) - (\ln(x) - 1)(1)}{(x - e)^2}$$ 6. **Simplify numerator:** $$\frac{1}{x}(x - e) = \frac{x - e}{x} = \cancel{\frac{x}{x}} - \frac{e}{x} = 1 - \frac{e}{x}$$ So numerator becomes: $$1 - \frac{e}{x} - \ln(x) + 1 = 2 - \ln(x) - \frac{e}{x}$$ 7. **Final derivative:** $$f'(x) = \frac{2 - \ln(x) - \frac{e}{x}}{(x - e)^2}$$ This is the derivative of the given function.