If f = 𝑥2 + 𝑦2 then 𝜕2𝑓 𝜕𝑥𝜕𝑦

To find the mixed partial derivative $\frac{\partial^2 f}{\partial x \partial y}$ of the function $f = x^2 + y^2$, let's follow these steps:

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

$f = x^2 + y^2$ Differentiate $f$ with respect to $x$ (treating $y$ as a constant): $\frac{\partial f}{\partial x} = 2x$

2. Second Partial Derivative with Respect to $y$:

Now, take the derivative of $\frac{\partial f}{\partial x} = 2x$ with respect to $y$ (treating $x$ as a constant): $\frac{\partial}{\partial y}(2x) = 0$ Since $2x$ is independent of $y$, its derivative with respect to $y$ is 0.

Final Answer:

$\frac{\partial^2 f}{\partial x \partial y} = 0$

Let me know if you'd like more details or have any further questions.

Related Questions:

1. What is the second partial derivative $\frac{\partial^2 f}{\partial y^2}$ for the same function?
2. How do mixed partial derivatives relate to the symmetry of a function?
3. How can we interpret the second mixed derivative in the context of geometry?
4. How would the result change if $f = x^2 + y^2 + xy$?
5. What is the condition for a function to have continuous mixed partial derivatives?

Tip:

When computing mixed partial derivatives, always differentiate one variable at a time and treat the others as constants.