1. **Problem:** Find the derivative and domain of the function $$f(x) = \sqrt{4 + \ln(x)}$$ and analyze related domain conditions for $$g(x) = \ln(x^2 - 12x)$$. 2. **Derivative of** $$f(x)$$: Given $$f(x) = \sqrt{4 + \ln(x)} = (4 + \ln(x))^{1/2}$$. Use the chain rule: $$f'(x) = \frac{1}{2}(4 + \ln(x))^{-1/2} \cdot \frac{d}{dx}(4 + \ln(x))$$ Since $$\frac{d}{dx}(4 + \ln(x)) = \frac{1}{x}$$, therefore, $$f'(x) = \frac{1}{2\sqrt{4 + \ln(x)}} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{4 + \ln(x)}}$$. 3. **Domain of** $$f(x)$$: The expression inside the square root must be non-negative: $$4 + \ln(x) \geq 0 \Rightarrow \ln(x) \geq -4$$ Exponentiate both sides: $$x \geq e^{-4}$$ Also, since $$\ln(x)$$ requires $$x > 0$$, final domain is: $$\boxed{[e^{-4}, \infty)}$$. 4. **Domain of** $$g(x) = \ln(x^2 - 12x)$$: Inside the logarithm, the argument must be positive: $$x^2 - 12x > 0$$ Factor: $$x(x - 12) > 0$$ Using sign analysis, - For $$x < 0$$, both factors $$x$$ and $$x - 12$$ have opposite signs: $$x < 0$$ and $$x - 12 < 0$$ gives positive product since negative * negative = positive. - For $$0 < x < 12$$ product is negative. - For $$x > 12$$ product is positive. Thus, domain of $$g(x)$$ is: $$(-\infty, 0) \cup (12, \infty)$$. 5. **Final answers:** - Derivative of $$f(x)$$: $$f'(x) = \frac{1}{2x\sqrt{4 + \ln(x)}}$$ - Domain of $$f(x)$$: $$[e^{-4}, \infty)$$ - Domain of $$g(x)$$: $$(-\infty, 0) \cup (12, \infty)$$