Given the function \( f(x) = x \ln x \), let's differentiate it using the **product rule** since it involves the product of two functions: \( x \) and \( \ln x \).

### Step-by-step differentiation:

1. The **product rule** states:  
   \[
   \frac{d}{dx}(u \cdot v) = u'v + uv'
   \]
   where \( u = x \) and \( v = \ln x \).

2. Now, differentiate each part:
   - \( u = x \) → \( u' = 1 \)
   - \( v = \ln x \) → \( v' = \frac{1}{x} \)

3. Apply the product rule:
   \[
   \frac{d}{dx} (x \ln x) = (1 \cdot \ln x) + (x \cdot \frac{1}{x})
   \]

4. Simplify:
   \[
   f'(x) = \ln x + 1
   \]

### Final Answer:
\[
f'(x) = \ln x + 1
\]

Would you like more details on any part of the process or have any questions?

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Here are 5 related questions you might want to explore:

1. What happens if \( f(x) = x^2 \ln x \)? How does the derivative change?
2. How would you compute the second derivative of \( f(x) = x \ln x \)?
3. What is the derivative of \( f(x) = x \ln(x^2) \)?
4. How would the derivative change if \( f(x) = \ln(x^2 + 1) \)?
5. How do we differentiate functions involving logarithms and powers like \( f(x) = x^3 \ln x \)?

**Tip:** The logarithmic differentiation technique is useful when dealing with functions of the form \( f(x) = x^n \ln x \), especially when \( n \) is a fraction or negative.

Explore the step-by-step differentiation of the function f(x) = x ln x using the product rule. Learn how to apply key calculus concepts, such as differentiating logarithmic functions and the product rule. This solution is ideal for high school or early university calculus students.

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