Problem 37 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the second-order partial derivatives of the function. In each case, show that the mixed partial derivatives \(f_{x y}\) and \(f_{y x}\) are equal. \(f(x, y)=x^{2}-2 x y+2 y^{2}+x-2 y\)

Step-by-Step Solution

Answer

The second-order partial derivatives of the function \(f(x, y) = x^{2} - 2xy + 2y^{2} + x - 2y\) are \(f_{xx} = 2\), \(f_{yy} = 4\), and \(f_{xy} = f_{yx} = -2\). Since the mixed partial derivatives, \(f_{xy}\) and \(f_{yx}\), are equal at -2, the requirement is satisfied.

Step 1: Find the first-order partial derivatives

To find the second-order partial derivatives, we must first find the first-order partial derivatives with respect to x and y. For the partial derivative with respect to x, differentiate the given function with respect to x, treating y as constant: \(f_x = \frac{\partial f}{\partial x} = 2x - 2y + 1\) For the partial derivative with respect to y, differentiate the given function with respect to y, treating x as constant: \(f_y = \frac{\partial f}{\partial y} = -2x + 4y - 2\)

Step 2: Find the second-order partial derivatives

Now, we will differentiate the first-order partial derivatives with respect to x and y to find the second-order partial derivatives. 1. Differentiate \(f_x\) with respect to x: \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2\) 2. Differentiate \(f_x\) with respect to y: \(f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -2\) 3. Differentiate \(f_y\) with respect to x: \(f_{yx} = \frac{\partial^2 f}{\partial y \partial x} = -2\) 4. Differentiate \(f_y\) with respect to y: \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4\)

Step 3: Show that the mixed partial derivatives are equal

From Step 2, we found that \(f_{xy} = -2\) and \(f_{yx} = -2\), which are equal. Therefore, the mixed partial derivatives, \(f_{xy}\) and \(f_{yx}\), are equal. The second-order partial derivatives are: - \(f_{xx} = 2\) - \(f_{yy} = 4\) - \(f_{xy} = f_{yx} = -2\)

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, crucial for understanding functions of several variables. To understand the partial derivative, imagine a function like a surface that depends on two variables, say x and y. Taking the partial derivative of this function with respect to x means figuring out how the function changes as x changes, while keeping y fixed as if y were a constant.

For instance, consider the function in our example, \(f(x, y)=x^{2}-2xy+2y^{2}+x-2y\). When we take the partial derivative of this function with respect to x, denoted by \(f_x\) or \(\frac{\partial f}{\partial x}\), we differentiate with respect to x and treat all other variables (in this case, just y) as constants. The result is \(2x - 2y + 1\). Similarly, the derivative of this function with respect to y, \(f_y\) or \(\frac{\partial f}{\partial y}\), yields \(-2x + 4y - 2\) when we treat x as constant.

Mixed Partial Derivatives

Mixed partial derivatives come into play when you take the derivative of a partial derivative. These are second-order partial derivatives, and they measure how the rate of change of the rate of change varies in different directions.

In our example, we first found \(f_x\) and \(f_y\). Then, we took the derivative of \(f_x\) with respect to y, yielding \(f_{xy} = -2\), and the derivative of \(f_y\) with respect to x, yielding \(f_{yx} = -2\). A key point in multivariable calculus is Clairaut's theorem which states that, under certain conditions, the order of differentiation does not matter—the mixed partial derivatives \(f_{xy}\) and \(f_{yx}\) will be equal. This is precisely what we observed in the example, which showcased that both mixed partial derivatives with respect to x then y, and y then x, are indeed the same.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of multiple variables. It entails the study of partial derivatives, multiple integrals, vector calculus, and other advanced mathematical concepts. These tools are vital for modeling and solving problems in physics, engineering, economics, and many other fields.

The function from the exercise, \(f(x, y)=x^{2}-2xy+2y^{2}+x-2y\), is a perfect example used to illustrate these advanced concepts. By analyzing its behavior through the lens of partial and mixed partial derivatives, we can better understand how it behaves in response to changes in different directions within a plane. Multivariable calculus helps us to not only find the slopes at a point in each independent direction but also understand how these slopes themselves change - a critical insight for fields like machine learning and optimization.