Final answer:
To differentiate the function f(x), apply the quotient rule and simplify. For the domain, we exclude the values where the denominator becomes zero and the natural log is undefined or negative, which results in the domain (5, e + 5) U (e + 5, infinity) in interval notation.
Step-by-step explanation:
To differentiate the function f(x) = \frac{x}{1 - \ln(x - 5)} and find its domain, we need to apply the quotient rule for differentiation and also determine when the denominator is not equal to zero, since division by zero is undefined.
Step 1: Differentiate f(x)
The quotient rule states if f(x) = \frac{g(x)}{h(x)}, then f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}. Here, g(x) = x and h(x) = 1 - \ln(x - 5). Thus, g'(x) = 1 and h'(x) = -\frac{1}{x - 5}. Applying the quotient rule gives us:
f'(x) = \frac{1 \cdot (1 - \ln(x - 5)) - x \cdot (-\frac{1}{x - 5})}{(1 - \ln(x - 5))^2} = \frac{1 - \ln(x - 5) + \frac{x}{x - 5}}{(1 - \ln(x - 5))^2}
Step 2: Find the domain of f(x)
The domain is all x-values where the function is defined. The restrictions on the domain come from the denominator, which cannot be zero and the natural logarithm, which requires positive arguments. Therefore, x - 5 > 0 and 1 - \ln(x - 5) \\eq 0. Solving these gives us the domain: x > 5 and \ln(x - 5) \\eq 1, or x \\eq e + 5. The general domain, considering these restrictions, is (5, e + 5) \cup (e + 5, \infty).