Problem 40 - Calculus Early Transcendentals

Question

Verify that \(f_{x y}=f_{y x}\) for the following functions. $$f(x, y)=\sqrt{x y}$$

Step-by-Step Solution

Answer

For the function \(f(x,y) = \sqrt{xy}\), the second partial derivatives are: $$f_{xy} = \frac{1}{2xy}$$ $$f_{yx} = \frac{1}{2xy}$$ Yes, the mixed partials are equal, confirming that \(f_{xy} = f_{yx}\).

Step 1: Find the partial derivative with respect to x (

The first step is to differentiate the function \(f(x,y) = \sqrt{xy}\) with respect to x while treating y as a constant. Using the chain rule for the square root function, we get: $$f_x = \frac{\partial}{\partial x} \sqrt{xy} = \frac{1}{2\sqrt{xy}} \frac{\partial}{\partial x} (xy)$$ Since \(\frac{\partial}{\partial x}(xy) = y\) (treating y as a constant), the partial derivative with respect to x is: $$f_x = \frac{y}{2\sqrt{xy}}$$

Step 2: Find the partial derivative with respect to y (

Now, we differentiate the function \(f(x,y) = \sqrt{xy}\) with respect to y, while treating x as a constant. Using the same method, we get: $$f_y = \frac{\partial}{\partial y} \sqrt{xy} = \frac{1}{2\sqrt{xy}} \frac{\partial}{\partial y} (xy)$$ Since \(\frac{\partial}{\partial y}(xy) = x\) (treating x as a constant), the partial derivative with respect to y is: $$f_y = \frac{x}{2\sqrt{xy}}$$

Step 3: Find the second partial derivatives (

Now, we differentiate \(f_x\) with respect to y (to obtain \(f_{xy}\)) and \(f_y\) with respect to x (to obtain \(f_{yx}\)). To get \(f_{xy}\), differentiate \(f_x\) with respect to y: $$f_{xy} = \frac{\partial}{\partial y} \left(\frac{y}{2\sqrt{xy}}\right) = \frac{1}{2}\frac{\partial}{\partial y} \left(\frac{y}{\sqrt{xy}}\right)$$ Taking the derivative, we find that: $$f_{xy} = \frac{1}{2}\left(\frac{\sqrt{xy}}{\sqrt{x^2y^2}}\right) = \frac{1}{2xy}$$ Next, we differentiate \(f_y\) with respect to x: $$f_{yx} = \frac{\partial}{\partial x} \left(\frac{x}{2\sqrt{xy}}\right) = \frac{1}{2}\frac{\partial}{\partial x} \left(\frac{x}{\sqrt{xy}}\right)$$ Taking the derivative, we find that: $$f_{yx} = \frac{1}{2}\left(\frac{\sqrt{xy}}{\sqrt{x^2y^2}}\right) = \frac{1}{2xy}$$

Step 4: Compare the second partial derivatives (

Comparing the second partial derivatives \(f_{xy}\) and \(f_{yx}\), we see that they are equal: $$f_{xy} = f_{yx} = \frac{1}{2xy}$$ This confirms that \(f_{xy} = f_{yx}\) for the function \(f(x,y) = \sqrt{xy}\).