The equation provided is $y = x^x + y^y$, and it asks to find the derivative $\frac{dy}{dx}$ for $x, y > 0$.

To solve this, we'll differentiate both sides of the equation with respect to $x$, using implicit differentiation since $y$ is a function of $x$. Let's proceed step-by-step:

1. Differentiate $y = x^x + y^y$ with respect to $x$:

   • For $x^x$, we use logarithmic differentiation: $\frac{d}{dx}(x^x) = x^x(\ln x + 1)$
   • For $y^y$, we apply implicit differentiation: $\frac{d}{dx}(y^y) = y^y(\ln y + 1)\frac{dy}{dx}$

2. Now, differentiate both sides: $\frac{dy}{dx} = \frac{d}{dx}(x^x) + \frac{d}{dx}(y^y)$ Substituting the differentiated forms: $\frac{dy}{dx} = x^x(\ln x + 1) + y^y(\ln y + 1)\frac{dy}{dx}$

3. Rearranging the equation to solve for $\frac{dy}{dx}$: $\frac{dy}{dx} - y^y(\ln y + 1)\frac{dy}{dx} = x^x(\ln x + 1)$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}\left(1 - y^y(\ln y + 1)\right) = x^x(\ln x + 1)$ Finally, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{x^x(\ln x + 1)}{1 - y^y(\ln y + 1)}$

This is the expression for $\frac{dy}{dx}$.

Would you like a detailed explanation of any step or further clarifications?

Here are some related questions for deeper understanding:

1. What is logarithmic differentiation, and why is it used for functions like $x^x$?
2. Can you explain how implicit differentiation is applied to functions where both sides involve the variable $y$?
3. How does the derivative of $y^y$ change if $y$ were independent of $x$?
4. What happens to the derivative expression when $y = 1$?
5. How would this process change if the equation involved powers of other variables (e.g., $y = x^z + y^z$)?

Tip: When differentiating powers involving the variable itself, logarithmic differentiation simplifies the process, allowing easier handling of such functions.