19. Verify that $f_{x y}=f_{y x}$ for the following function. \[ \begin{array}{l} f(x, y)=2 \cos x y \\ f_{x}=\square \\ f_{y}=\square \\ f_{x y}=f_{y x}=\square \end{array} \]
Solution
Step 1 ：Given the function \(f(x, y)=2 \cos(xy)\).
Step 2 ：Find the first partial derivatives \(f_x\) and \(f_y\). The first partial derivative of \(f\) with respect to \(x\) is \(f_x = -2y\sin(xy)\) and with respect to \(y\) is \(f_y = -2x\sin(xy)\).
Step 3 ：Find the second partial derivatives \(f_{xy}\) and \(f_{yx}\). The second partial derivative of \(f\) with respect to \(x\) then \(y\) is \(f_{xy} = -2xy\cos(xy) - 2\sin(xy)\) and with respect to \(y\) then \(x\) is \(f_{yx} = -2xy\cos(xy) - 2\sin(xy)\).
Step 4 ：Compare \(f_{xy}\) and \(f_{yx}\) to see if they are equal. Since \(f_{xy} = f_{yx} = -2xy\cos(xy) - 2\sin(xy)\), the function \(f(x, y)=2 \cos(xy)\) satisfies the equality of mixed partials.
Step 5 ：\(\boxed{f_{xy}=f_{yx}=-2xy\cos(xy) - 2\sin(xy)}\) is the final answer.
