To find the derivative of the function $f(x) = x^x$, we can use logarithmic differentiation, as the variable $x$ appears in both the base and the exponent.

Steps:

1. Rewrite the Function Using Logarithms:

   Take the natural logarithm of both sides: $y = x^x$ $\ln y = \ln(x^x)$ Simplify using the property $\ln(a^b) = b \ln a$: $\ln y = x \ln x$

2. Differentiate Both Sides with Respect to $x$:

   Applying the derivative to both sides, using the chain rule on the left: $\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx} (x \ln x)$

   Now, differentiate $x \ln x$ using the product rule: $\frac{d}{dx} (x \ln x) = \ln x + 1$

   So: $\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + 1$

3. Solve for $\frac{dy}{dx}$:

   Substitute $y = x^x$ back into the equation: $\frac{dy}{dx} = x^x (\ln x + 1)$

Final Answer:

$\frac{d}{dx} (x^x) = x^x (\ln x + 1)$

Would you like more details on any of these steps?

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Here are some related questions for further exploration:

1. How would you find the second derivative of $x^x$?
2. Can you apply logarithmic differentiation to $x^{x^x}$?
3. What happens if $x$ is a complex number in $x^x$?
4. How does this derivative compare to $a^x$, where $a$ is a constant?
5. How would you integrate $x^x$?

Tip: Logarithmic differentiation is especially useful for differentiating functions where variables are both in the base and the exponent!