Question

Verify Fxy=Fyx for f(x, y) =in(xy2)

Answer 1

To verify that \( F_{xy} = F_{yx} \) for the function \( f(x, y) = \ln(xy^2) \), we need to calculate the mixed partial derivatives \( F_{xy} \) and \( F_{yx} \).

### Step 1: Calculate the first partial derivatives

1. **First partial derivative with respect to \( x \)**:

\[
f(x, y) = \ln(xy^2) = \ln(x) + \ln(y^2) = \ln(x) + 2\ln(y)
\]

Now, differentiate with respect to \( x \):

\[
f_x = \frac{\partial}{\partial x} \ln(xy^2) = \frac{\partial}{\partial x} \ln(x) + \frac{\partial}{\partial x} 2\ln(y) = \frac{1}{x} + 0 = \frac{1}{x}
\]

2. **First partial derivative with respect to \( y \)**:

Now, differentiate \( f(x, y) \) with respect to \( y \):

\[
f_y = \frac{\partial}{\partial y} \ln(xy^2) = \frac{\partial}{\partial y} \ln(x) + \frac{\partial}{\partial y} 2\ln(y) = 0 + \frac{2}{y} = \frac{2}{y}
\]

### Step 2: Calculate the mixed partial derivatives

1. **Mixed partial derivative \( F_{xy} \)**:

Differentiate \( f_x \) with respect to \( y \):

\[
f_{xy} = \frac{\partial}{\partial y} \left( \frac{1}{x} \right) = 0
\]

2. **Mixed partial derivative \( F_{yx} \)**:

Differentiate \( f_y \) with respect to \( x \):

\[
f_{yx} = \frac{\partial}{\partial x} \left( \frac{2}{y} \right) = 0
\]

### Step 3: Compare the mixed partial derivatives

Now we compare the values:

\[
f_{xy} = 0 \quad \text{and} \quad f_{yx} = 0
\]

### Conclusion

Since \( f_{xy} = f_{yx} \), we have verified that the mixed partial derivatives are equal:

\[
F_{xy} = F_{yx}
\]

Thus, the equality \( F_{xy} = F_{yx} \) holds true for the function \( f(x, y) = \ln(xy^2) \).