Differentiate \( f \) and find the domain of \( f \). (Enter the domain in interval notation.) \[ f(x)=\ln \left(x^{2}-18 x\right) \] derivative \( f^{\prime}(x)=\square \) domain

We begin with the function \[ f(x)=\ln\left(x^2-18x\right). \] **Step 1. Find the derivative.** The derivative of \( f \) is found using the chain rule. Let \[ u(x)=x^2-18x. \] Then \[ f(x)=\ln\left(u(x)\right) \quad \text{and} \quad f'(x)=\frac{u'(x)}{u(x)}. \] Differentiate \( u(x) \): \[ u'(x)=\frac{d}{dx}\left(x^2-18x\right)=2x-18. \] Thus, the derivative is \[ f'(x)=\frac{2x-18}{x^2-18x}. \] This expression can be factored further by noting that \[ 2x-18=2(x-9) \quad \text{and} \quad x^2-18x=x(x-18). \] So, an equivalent form is \[ f'(x)=\frac{2(x-9)}{x(x-18)}. \] **Step 2. Determine the Domain.** The function \( f(x)=\ln\left(x^2-18x\right) \) is defined when the argument of the logarithm is positive: \[ x^2-18x>0. \] Factor the quadratic: \[ x^2-18x=x(x-18). \] We now solve \[ x(x-18)>0. \] Identify the critical points: \( x=0 \) and \( x=18 \). We test intervals defined by these points. - For \( x<0 \), choose \( x=-1 \): \[ (-1)(-1-18)=(-1)(-19)=19>0. \] - For \( 018 \), choose \( x=20 \): \[ (20)(20-18)=20(2)=40>0. \] Thus, the inequality holds for: \[ x\in (-\infty,0) \cup (18,\infty). \] **Final Answers:** Derivative: \[ f'(x)=\frac{2x-18}{x^2-18x}=\frac{2(x-9)}{x(x-18)}. \] Domain: \[ (-\infty,0) \cup (18,\infty). \]