The problem provided asks for the derivative of the function $f(x) = x^x$.

Solution:

The function $f(x) = x^x$ can be rewritten using logarithms to facilitate differentiation.

Step 1: Rewrite the function

Using the property of logarithms: $x^x = e^{x \ln(x)}.$ So, the function becomes: $f(x) = e^{x \ln(x)}.$

Step 2: Differentiate

Now, differentiate $f(x)$ with respect to $x$. Using the chain rule: $f'(x) = e^{x \ln(x)} \cdot \frac{d}{dx}(x \ln(x)).$

Step 3: Differentiate $x \ln(x)$

The derivative of $x \ln(x)$ is: $\frac{d}{dx}(x \ln(x)) = \ln(x) + 1.$

Step 4: Combine results

Substitute this back: $f'(x) = e^{x \ln(x)} \cdot (\ln(x) + 1).$

Since $e^{x \ln(x)} = x^x$, the derivative simplifies to: $f'(x) = x^x \cdot (\ln(x) + 1).$

Final Answer:

$f'(x) = x^x (\ln(x) + 1).$

Let me know if you'd like additional details or examples!

Related Questions:

1. How do logarithmic properties simplify differentiation of exponential functions?
2. What are the applications of $f(x) = x^x$ in real-world problems?
3. How do you apply the chain rule in other exponential-logarithmic expressions?
4. Can this derivative be extended to fractional or negative values of $x$?
5. How does $f(x) = x^x$ behave as $x \to 0^+$ or $x \to \infty$?

Tip:

When differentiating functions involving $x^x$, always start by rewriting them with the exponential property $x^x = e^{x \ln(x)}$ to simplify the process.