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

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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:
$$
(-\infty,0) \cup (18,\infty).
$$
Derivative: $$ f'(x) = \frac{2(x - 9)}{x(x - 18)} $$ Domain: $$ (-\infty, 0) \cup (18, \infty) $$

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

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