To find $\frac{\partial^2 u}{\partial x \partial y}$ for $u = x^y + y^x$, let's compute the derivatives step by step.

Step 1: First Partial Derivative with Respect to $x$

Given $u = x^y + y^x$, differentiate with respect to $x$: $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^y) + \frac{\partial}{\partial x}(y^x).$

1. For $x^y$, treat $y$ as a constant: $\frac{\partial}{\partial x}(x^y) = yx^{y-1}.$

2. For $y^x$, apply the chain rule since $y^x = e^{x \ln y}$: $\frac{\partial}{\partial x}(y^x) = y^x \ln y.$

Thus: $\frac{\partial u}{\partial x} = yx^{y-1} + y^x \ln y.$

Step 2: Second Mixed Partial Derivative ($\frac{\partial^2 u}{\partial x \partial y}$)

Now, differentiate $\frac{\partial u}{\partial x}$ with respect to $y$: $\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial}{\partial y}\left(yx^{y-1} + y^x \ln y\right).$

1. Differentiate $yx^{y-1}$ with respect to $y$:

   • Product rule: $y \cdot x^{y-1}$: $\frac{\partial}{\partial y}(yx^{y-1}) = x^{y-1} + y \cdot x^{y-1} \ln x.$

2. Differentiate $y^x \ln y$ with respect to $y$:

   • Apply the product rule: $\frac{\partial}{\partial y}(y^x \ln y) = \frac{\partial}{\partial y}(y^x) \ln y + y^x \cdot \frac{\partial}{\partial y}(\ln y).$
   • For $\frac{\partial}{\partial y}(y^x)$, treat $x$ as constant: $\frac{\partial}{\partial y}(y^x) = xy^{x-1}.$
   • For $\frac{\partial}{\partial y}(\ln y)$: $\frac{\partial}{\partial y}(\ln y) = \frac{1}{y}.$ Thus: $\frac{\partial}{\partial y}(y^x \ln y) = xy^{x-1} \ln y + \frac{y^x}{y}.$

Combine the results: $\frac{\partial^2 u}{\partial x \partial y} = x^{y-1} + yx^{y-1} \ln x + xy^{x-1} \ln y + \frac{y^x}{y}.$

Final Answer:

$\frac{\partial^2 u}{\partial x \partial y} = x^{y-1} + yx^{y-1} \ln x + xy^{x-1} \ln y + y^{x-1}.$

Let me know if you'd like any clarification or further assistance!

Here are five related questions to deepen understanding:

1. How would you compute $\frac{\partial^2 u}{\partial y \partial x}$, and would it yield the same result?
2. What happens to $\frac{\partial^2 u}{\partial x \partial y}$ if $x = y$?
3. Can you generalize this process to a different function like $u = x^a + y^b$?
4. What is the physical interpretation of mixed partial derivatives in this context?
5. How do logarithmic and exponential rules simplify such derivative calculations?

Tip: Always check if mixed partial derivatives are equal ($\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial^2 u}{\partial y \partial x}$) by Clairaut's theorem for consistency!