Differentiate $$ f $$ and find the domain of $$ f $$. (Enter the domain in interval notation.) $$ f(x)=\frac{x}{1-\ln (x-3)} $$ derivative $$ f^{\prime}(x)=\square $$ domain

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We are given 
$$
f(x)=\frac{x}{1-\ln(x-3)}.
$$
Below are the steps to differentiate $$ f(x) $$ and find its domain.

### 1. Domain of $$ f(x) $$
The function involves a natural logarithm, so its argument must be positive:
$$
x-3 > 0 \quad \Longrightarrow \quad x > 3.
$$
Also, the denominator must not be zero:
$$
1-\ln(x-3) \neq 0 \quad \Longrightarrow \quad \ln(x-3) \neq 1 \quad \Longrightarrow \quad x-3 \neq e \quad \Longrightarrow \quad x \neq 3+e.
$$
Thus, the domain in interval notation is:
$$
(3,\, 3+e) \cup (3+e,\, \infty).
$$

### 2. Differentiation of $$ f(x) $$
We use the quotient rule. For a function of the form $$\frac{u(x)}{v(x)}$$,
$$
f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}.
$$

Let 
$$
u(x)=x \quad \text{and} \quad v(x)=1-\ln(x-3).
$$
Compute their derivatives:
- $$ u'(x)=1 $$.
- To differentiate $$ v(x) $$, note that
  $$
  v'(x)=-\frac{d}{dx} \ln(x-3)=-\frac{1}{x-3}.
  $$

Now apply the quotient rule:
$$
f'(x)=\frac{1\cdot \left(1-\ln(x-3)\right)-x\left(-\frac{1}{x-3}\right)}{\left(1-\ln(x-3)\right)^2}.
$$
Simplify the numerator:
$$
1-\ln(x-3) + \frac{x}{x-3}.
$$
Thus, the derivative is:
$$
f'(x)=\frac{1-\ln(x-3)+\frac{x}{x-3}}{\left(1-\ln(x-3)\right)^2}.
$$

### Final Answer
- Derivative:
  $$
  f^{\prime}(x)=\frac{1-\ln(x-3)+\frac{x}{x-3}}{\left(1-\ln(x-3)\right)^2}.
  $$
- Domain:
  $$
  (3,\, 3+e) \cup (3+e,\, \infty).
  $$
Derivative: $$ f^{\prime}(x)=\frac{1-\ln(x-3)+\frac{x}{x-3}}{\left(1-\ln(x-3)\right)^2} $$ Domain: $$ (3,\, 3+e) \cup (3+e,\, \infty) $$

Differentiate and find the domain of . (Enter the domain in interval notation.)

derivative
domain

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Differentiate and find the domain of . (Enter the domain in interval notation.) derivative domain