I can see the mathematical content in your uploaded photo. Let's solve the problem step-by-step.

The task is to find the derivative of $y = x^x$, and you are given multiple-choice answers.

Step 1: Rewrite the function

The function $y = x^x$ can be rewritten using logarithms to simplify differentiation: $y = x^x = e^{x \ln x}.$

Step 2: Differentiate $y = e^{x \ln x}$

Using the chain rule, the derivative of $e^{u}$ is: $\frac{dy}{dx} = e^u \cdot \frac{du}{dx},$ where $u = x \ln x$.

Step 3: Differentiate $u = x \ln x$

Using the product rule for $u = x \ln x$: $\frac{du}{dx} = \ln x + 1.$

Step 4: Substitute $\frac{du}{dx}$ into $\frac{dy}{dx}$

Now substitute $u = x \ln x$ and $\frac{du}{dx} = \ln x + 1$ into the derivative formula: $\frac{dy}{dx} = e^{x \ln x} \cdot (\ln x + 1).$

Since $e^{x \ln x} = x^x$, the final derivative is: $\frac{dy}{dx} = x^x (\ln x + 1).$

Step 5: Match with the provided options

The correct answer is: $\boxed{x^x (\ln x + 1)}.$

Would you like me to explain any steps in more detail?

5 Related Questions:

1. What is the product rule and how is it applied in differentiation?
2. How does the chain rule work in composite functions?
3. Why is logarithmic differentiation useful for functions like $x^x$?
4. Can you derive the formula for $x^a$ (where $a$ is a constant)?
5. How can exponential functions be simplified during differentiation?

Tip:

When differentiating functions with variable bases and exponents, rewrite the function using natural logarithms to simplify the process.