Suppose f(x,y) and all higher partial derivatives are continuous functions, and fyxy(x,y)=2xy2. What is the value of fxyy(2,1)?


Given that \( f(x, y) \) and all higher partial derivatives are continuous functions, and that \( f_{yxy}(x, y) = 2xy^2 \), we want to find the value of \( f_{xyy}(2, 1) \).

Using the symmetry of mixed partial derivatives (Clairaut's theorem), which states that if all partial derivatives are continuous, the order of differentiation does not matter:

\[
f_{yxy}(x, y) = f_{yyx}(x, y) = f_{xyy}(x, y).
\]

Given this property, we have:

\[
f_{xyy}(x, y) = f_{yxy}(x, y).
\]

So,

\[
f_{xyy}(2, 1) = f_{yxy}(2, 1).
\]

Substitute \( x = 2 \) and \( y = 1 \) into \( f_{yxy}(x, y) = 2xy^2 \):

\[
f_{yxy}(2, 1) = 2 \times 2 \times 1^2 = 4.
\]

Thus, the value of \( f_{xyy}(2, 1) \) is \( 4 \).

### Summary
The value of \( f_{xyy}(2, 1) \) is **4**.

Would you like a more detailed explanation or have any other questions?

### Related Questions:
1. What is Clairaut's theorem and how is it used in mixed partial derivatives?
2. How do you find higher-order partial derivatives for functions of two variables?
3. What conditions must be met for the mixed partial derivatives to be equal?
4. Can you provide an example where the order of partial differentiation matters?
5. How does the continuity of a function affect its partial derivatives?

### Tip:
When working with partial derivatives, always verify the continuity of the function and its partial derivatives to apply the symmetry of mixed partials correctly.

Suppose f(x,y) and all higher partial derivatives are continuous functions, and fyxy(x,y)=2xy^2. What is the value of fxyy(2,1)?

Solve for fxyy(2, 1) given f(x,y) and continuous higher partial derivatives with fyxy(x,y) = 2xy^2. This problem uses Clairaut's Theorem and covers concepts of mixed partial derivatives. Suitable for undergraduate calculus students.

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