Let's solve the given problem step by step.

The question is asking for the mixed partial derivative $\frac{\partial^2 f(x, y)}{\partial x \partial y}$ for $f(x, y) = e^{xy}$.

Solution:

1. First Partial Derivative with respect to $x$: Start by taking the partial derivative of $f(x, y) = e^{xy}$ with respect to $x$, treating $y$ as a constant: $\frac{\partial f(x, y)}{\partial x} = \frac{\partial}{\partial x}(e^{xy}) = y \cdot e^{xy}.$

2. Second Partial Derivative with respect to $y$: Now, take the partial derivative of $\frac{\partial f(x, y)}{\partial x} = y \cdot e^{xy}$ with respect to $y$: $\frac{\partial^2 f(x, y)}{\partial x \partial y} = \frac{\partial}{\partial y}(y \cdot e^{xy}).$

   Using the product rule ($\frac{d}{dy}(uv) = u'v + uv'$): $\frac{\partial^2 f(x, y)}{\partial x \partial y} = \frac{\partial}{\partial y}(y) \cdot e^{xy} + y \cdot \frac{\partial}{\partial y}(e^{xy}).$

   • The derivative of $y$ with respect to $y$ is $1$, so: $\frac{\partial}{\partial y}(y) \cdot e^{xy} = e^{xy}.$
   • The derivative of $e^{xy}$ with respect to $y$ is $x \cdot e^{xy}$, so: $y \cdot \frac{\partial}{\partial y}(e^{xy}) = y \cdot x \cdot e^{xy} = xye^{xy}.$

   Adding these two terms: $\frac{\partial^2 f(x, y)}{\partial x \partial y} = e^{xy} + xye^{xy}.$

Final Answer:

$\boxed{e^{xy} + xye^{xy}}$

Let me know if you'd like a more detailed explanation or further clarifications! 😊

Related Questions:

1. How would the result differ if $f(x, y) = e^{x^2y}$?
2. What is the second mixed partial derivative $\frac{\partial^2 f}{\partial y \partial x}$? Is it equal to $\frac{\partial^2 f}{\partial x \partial y}$?
3. Can you find all second-order partial derivatives of $f(x, y) = e^{xy}$?
4. What does Clairaut's theorem say about the equality of mixed partial derivatives?
5. How do mixed partial derivatives change for trigonometric functions like $f(x, y) = \sin(xy)$?

Tip:

When calculating mixed partial derivatives, always remember to treat the variable you're not differentiating with respect to as a constant.