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.