Verify that \( f_{x y}=f_{y x} \), for the function \( f(x, y)=4 x^{5}+3 y^{7}+2 \). For the function \( f(x, y)=4 x^{5}+3 y^{7}+2, f_{x y}=f_{y x} \) since \( f_{x}=\square \) and \( f_{y}=\square \) Therefore, \( f_{x y}=\square \) and \( f_{y x}=\square \)

To verify that \( f_{xy} = f_{yx} \), we first need to find the first-order partial derivatives. The partial derivative of \( f \) with respect to \( x \) is \( f_x = \frac{\partial}{\partial x}(4x^5 + 3y^7 + 2) = 20x^4 \). Next, taking the partial derivative of \( f_x \) with respect to \( y \) gives \( f_{xy} = \frac{\partial}{\partial y}(20x^4) = 0 \) (since \( 20x^4 \) is independent of \( y \)). Now, let's find \( f_y \). The partial derivative of \( f \) with respect to \( y \) is \( f_y = \frac{\partial}{\partial y}(4x^5 + 3y^7 + 2) = 21y^6 \). Taking the partial derivative of \( f_y \) with respect to \( x \) gives \( f_{yx} = \frac{\partial}{\partial x}(21y^6) = 0 \) (as \( 21y^6 \) is independent of \( x \)). Thus, we have \( f_{xy} = 0 \) and \( f_{yx} = 0 \), meaning \( f_{xy} = f_{yx} \) holds true. So, to fill in the squares: \( f_x = 20x^4 \) and \( f_y = 21y^6 \), therefore \( f_{xy} = 0 \) and \( f_{yx} = 0 \).