Problem 36 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Verify that the second-order mixed partial derivatives of the function \(f(x, y) = xe^y\) are equal. Solution: Step 1: Find the first-order partial derivatives, where \(f_x = e^y\) and \(f_y = xe^y\). Step 2: Find the second-order mixed partial derivatives, where \(f_{xy} = e^y\) and \(f_{yx} = e^y\). Step 3: Compare the second-order mixed partial derivatives, since \(f_{xy} = e^y = f_{yx}\), they are equal. Thus, the second-order mixed partial derivatives of the function \(f(x, y) = xe^y\) are equal.

Step 1: Find the first-order partial derivatives

To find the first-order partial derivative with respect to x, we will treat y as a constant and differentiate the function with respect to x: $$f_x = \frac{\partial}{\partial x}(xe^y)$$ Similarly, to find the first-order partial derivative with respect to y, we will treat x as a constant and differentiate the function with respect to y: $$f_y = \frac{\partial}{\partial y}(xe^y)$$ Computing these derivatives, we get: $$f_x = e^y$$ $$f_y = xe^y$$

Step 2: Find the second-order mixed partial derivatives

Now, we'll find the second-order mixed partial derivatives by taking the derivative of the first-order derivatives with respect to the other variable. To find \(f_{xy}\), differentiate \(f_x\) with respect to y, treating x as a constant: $$f_{xy} = \frac{\partial}{\partial y}(e^y)$$ To find \(f_{yx}\), differentiate \(f_y\) with respect to x, treating y as a constant: $$f_{yx} = \frac{\partial}{\partial x}(xe^y)$$ Computing these derivatives, we get: $$f_{xy} = e^y$$ $$f_{yx} = e^y$$

Step 3: Compare the second-order mixed partial derivatives

Now, we will compare \(f_{xy}\) and \(f_{yx}\) to see if they are equal. We found that both of these derivatives equal \(e^y\): $$f_{xy} = e^y = f_{yx}$$ Since \(f_{xy} = f_{yx}\), this verifies that the second-order mixed partial derivatives of the function \(f(x, y) = xe^y\) are indeed equal.