91Ó°ÊÓ

First, we need to find the partial derivatives of the function with respect to \(x\) and \(y\). That is, we need to find \(f_x\) and \(f_y\). Using the chain rule, we obtain: $$f_x = \dfrac{\partial}{\partial x}\sqrt{xy} = \dfrac{1}{2\sqrt{xy}} (\partial (xy) / \partial x)$$ and $$f_y = \dfrac{\partial}{\partial y}\sqrt{xy} = \dfrac{1}{2\sqrt{xy}} (\partial (xy) / \partial y)$$ Now we can calculate the partial derivatives inside the parentheses: $$\dfrac{\partial(xy)}{\partial x} = y$$ and $$\dfrac{\partial(xy)}{\partial y} = x$$ So we have: $$f_x = \dfrac{1}{2\sqrt{xy}} (y) = \dfrac{y}{2\sqrt{xy}}$$ and $$f_y = \dfrac{1}{2\sqrt{xy}} (x) = \dfrac{x}{2\sqrt{xy}}$$

Now that we have \(f_x\) and \(f_y\), we can find the mixed second partial derivatives \(f_{xy}\) and \(f_{yx}\). This means calculating the partial derivative of \(f_x\) with respect to \(y\) and the partial derivative of \(f_y\) with respect to \(x\): $$f_{xy} = \dfrac{\partial}{\partial y} \left(\dfrac{y}{2\sqrt{xy}}\right)$$ and $$f_{yx} = \dfrac{\partial}{\partial x} \left(\dfrac{x}{2\sqrt{xy}}\right)$$ We will calculate these derivatives using the quotient and chain rules: For \(f_{xy}\): $$f_{xy} = \dfrac{2\sqrt{xy} \cdot 1 - y \cdot \dfrac{1}{2} \cdot (xy)^{-1/2} \cdot x}{ (2\sqrt{xy})^2}$$ $$f_{xy} = \dfrac{2\sqrt{xy} - \dfrac{xy}{2\sqrt{xy}}}{4xy}$$ $$f_{xy} = \dfrac{\dfrac{4xy}{2\sqrt{xy}} - \dfrac{xy}{2\sqrt{xy}}}{4xy}$$ $$f_{xy} = \dfrac{3xy}{4xy\sqrt{xy}} = \dfrac{3}{4\sqrt{xy}}$$ For \(f_{yx}\): $$f_{yx} = \dfrac{2\sqrt{xy} \cdot 1 - x \cdot \dfrac{1}{2} \cdot (xy)^{-1/2} \cdot y}{ (2\sqrt{xy})^2}$$ $$f_{yx} = \dfrac{2\sqrt{xy} - \dfrac{xy}{2\sqrt{xy}}}{4xy}$$ $$f_{yx} = \dfrac{\dfrac{4xy}{2\sqrt{xy}} - \dfrac{xy}{2\sqrt{xy}}}{4xy}$$ $$f_{yx} = \dfrac{3xy}{4xy\sqrt{xy}} = \dfrac{3}{4\sqrt{xy}}$$

Comparing the mixed second partial derivatives we found in Step 2, we can see that: $$f_{xy} = \dfrac{3}{4\sqrt{xy}} = f_{yx}$$ Therefore, it is verified that \(f_{xy} = f_{yx}\) for the given function \(f(x, y) = \sqrt{xy}\).