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The task is to show that the mixed partial derivatives are equal: \( f_{xyy} = f_{yyx} = f_{yxy} \), for the function \( f(x, y) = \sin(xy) \).

### Steps:
1. **Find the first partial derivatives**:
   - The function is \( f(x, y) = \sin(xy) \).
   - First, compute the partial derivatives with respect to \( x \) and \( y \).

   \[
   f_x = \frac{\partial}{\partial x} \sin(xy) = y \cos(xy)
   \]
   \[
   f_y = \frac{\partial}{\partial y} \sin(xy) = x \cos(xy)
   \]

2. **Find the second partial derivatives**:
   - Compute the second partial derivative with respect to \( y \) first.

   \[
   f_{yy} = \frac{\partial}{\partial y} \left( x \cos(xy) \right) = x \left( -x \sin(xy) \right) = -x^2 \sin(xy)
   \]

   - Now compute the mixed partial derivative \( f_{xy} \).

   \[
   f_{xy} = \frac{\partial}{\partial y} \left( y \cos(xy) \right) = \cos(xy) - xy \sin(xy)
   \]

3. **Find the third mixed partial derivatives**:
   - Compute \( f_{xyy} \), \( f_{yxy} \), and \( f_{yyx} \) and show they are equal.

   \[
   f_{xyy} = \frac{\partial}{\partial y} \left( \cos(xy) - xy \sin(xy) \right) = -x \sin(xy) - x \sin(xy) - xy^2 \cos(xy)
   \]
   \[
   f_{yxy} = \frac{\partial}{\partial x} \left( -x^2 \sin(xy) \right) = -x^2 \cos(xy) + x^3 \sin(xy)
   \]
   \[
   f_{yyx} = \frac{\partial}{\partial x} \left( -x^3

Show that f_xyy = f_yxy = f_yyx if f(x, y) = sin(xy).

Learn how to prove the equality of mixed partial derivatives for the function f(x, y) = sin(xy). This problem involves calculating multiple derivatives and applying Schwarz's Theorem. Suitable for university students in advanced calculus.

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