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The problem asks to show that \( f_{xyy} = f_{yyx} = f_{yxy} \), where the function is given as:

\[
f(x, y) = \sin(x + y)
\]

### Solution Steps

1. **Find the first partial derivatives:**

   First, differentiate \( f(x, y) = \sin(x + y) \) with respect to \( x \) and \( y \).

   - \( f_x = \frac{\partial}{\partial x} \sin(x + y) = \cos(x + y) \)
   - \( f_y = \frac{\partial}{\partial y} \sin(x + y) = \cos(x + y) \)

2. **Find the second partial derivatives:**

   Now, compute the second derivatives by differentiating again with respect to \( x \) and \( y \).

   - \( f_{xx} = \frac{\partial}{\partial x} \left( \cos(x + y) \right) = -\sin(x + y) \)
   - \( f_{yy} = \frac{\partial}{\partial y} \left( \cos(x + y) \right) = -\sin(x + y) \)
   - \( f_{xy} = \frac{\partial}{\partial y} \left( \cos(x + y) \right) = -\sin(x + y) \)
   - \( f_{yx} = \frac{\partial}{\partial x} \left( \cos(x + y) \right) = -\sin(x + y) \)

   So we see that \( f_{xy} = f_{yx} \).

3. **Find the third partial derivatives:**

   Next, compute the third partial derivatives.

   - \( f_{xyy} = \frac{\partial}{\partial y} \left( f_{xy} \right) = \frac{\partial}{\partial y} \left( -\sin(x + y) \right) = -\cos(x + y) \)
   - \( f_{yyx} = \frac{\partial}{\partial x} \left( f_{yy} \right) = \frac{\partial}{\partial x} \left( -\sin(x + y) \right) = -\cos(x + y) \)
   - \( f_{yxy} = \frac{\partial}{\partial y} \left( f_{yx} \right) = \frac{\partial}{\partial y} \left( -\sin(x + y) \right) = -\cos(x + y) \)

   Therefore, we have shown that:
   \[
   f_{xyy} = f_{yyx} = f_{yxy} = -\cos(x + y)
   \]

### Conclusion:
Hence, \( f_{xyy} = f_{yyx} = f_{yxy} \) as required.

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Would you like more details or further clarifications? Here are five related questions for deeper understanding:

1. How would the result change if the function was \( f(x, y) = \sin(xy) \)?
2. What is the geometric interpretation of mixed partial derivatives being equal?
3. Can you prove the equality of mixed partials using Clairaut's theorem?
4. What would happen if the function was not continuously differentiable?
5. How does this result extend to higher dimensions or more complex functions?

**Tip**: When computing partial derivatives of trigonometric functions, remember the chain rule is essential in handling composite expressions like \( \sin(x + y) \).

Show that f_{xyy} = f_{yyx} = f_{yxy} if f(x, y) = sin(x + y)

Learn how to prove that f_{xyy} = f_{yyx} = f_{yxy} for the function f(x, y) = sin(x + y) using partial derivatives and Clairaut's theorem. Detailed step-by-step solution for undergraduate calculus students.

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