Problem 21 - Applied Calculus

Question

Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. $$ f(x, y)=x^{2}+2 x y+y^{2} $$

Step-by-Step Solution

Answer

The second-order partial derivatives are \(f_{xx} = 2\), \(f_{yy} = 2\), \(f_{xy} = 2\), \(f_{yx} = 2\); mixed partials are equal.

Step 1: Find the First Partial Derivative with respect to x

To find the first partial derivative of \(f(x, y)\) with respect to \(x\), differentiate each term of the function treating \(y\) as a constant. \[ f_x(x, y) = \frac{\partial}{\partial x}(x^2 + 2xy + y^2) = 2x + 2y \]

Step 2: Find the First Partial Derivative with respect to y

Now differentiate \(f(x, y)\) with respect to \(y\), treating \(x\) as a constant. \[ f_y(x, y) = \frac{\partial}{\partial y}(x^2 + 2xy + y^2) = 2x + 2y \]

Step 3: Find the Second Partial Derivative \(f_{xx}\)

Differentiate \(f_x(x, y) = 2x + 2y\) with respect to \(x\). \[ f_{xx}(x, y) = \frac{\partial}{\partial x}(2x + 2y) = 2 \]

Step 4: Find the Second Partial Derivative \(f_{yy}\)

Differentiate \(f_y(x, y) = 2x + 2y\) with respect to \(y\). \[ f_{yy}(x, y) = \frac{\partial}{\partial y}(2x + 2y) = 2 \]

Step 5: Find the Mixed Partial Derivative \(f_{xy}\)

First, we differentiate \(f_x(x, y) = 2x + 2y\) with respect to \(y\). \[ f_{xy}(x, y) = \frac{\partial}{\partial y}(2x + 2y) = 2 \]

Step 6: Find the Mixed Partial Derivative \(f_{yx}\)

Now differentiate \(f_y(x, y) = 2x + 2y\) with respect to \(x\). \[ f_{yx}(x, y) = \frac{\partial}{\partial x}(2x + 2y) = 2 \]

Step 7: Confirm Equality of Mixed Partials

The mixed partial derivatives obtained are \(f_{xy} = 2\) and \(f_{yx} = 2\). Since \(f_{xy} = f_{yx}\), the mixed partials are indeed equal.

Key Concepts

Second-Order Derivatives

Second-order derivatives are essentially the derivatives of derivatives. Think of it as taking a further step to see how the initial rate of change itself is changing. In mathematics, we often use this concept to examine the curvature or concavity of a function. For example, when you calculate the second partial derivative, such as \( f_{xx} \) or \( f_{yy} \), you are checking how the rate of change with respect to \( x \) or \( y \) changes again in the direction of \( x \) or \( y \) respectively.
This can provide insight into the shape of the graph of a function:

• Positive second-order derivatives typically indicate a portion of the graph that is curving upwards.
• Negative second-order derivatives suggest the graph is curving downwards.

In this exercise, both \( f_{xx}(x, y) = 2 \) and \( f_{yy}(x, y) = 2 \) are positive, indicating that the function curves upwards, meaning it has a surface that bowls upwards when visualizing in three dimensions.

Mixed Partial Derivatives

Mixed partial derivatives arise when you take the derivative of a function first with respect to one variable and then with respect to another variable. This might sound a bit complex at first, but it's quite manageable.
When we look at \( f_{xy} \) or \( f_{yx} \), we are finding out how the derivative with respect to one variable alters when we vary another. It answers the question: "how does the change in \( x \) affect the change in \( y \)?" or vice versa. This is deeply rooted in calculus’ multidimensional analysis.

• \( f_{xy} \) involves differentiating partially with respect to \( x \) first, and then with respect to \( y \).
• \( f_{yx} \) involves differentiating partially with respect to \( y \) first, and then with respect to \( x \).

The symmetry of these derivatives, demonstrated by \( f_{xy} = f_{yx} \), is a result of Clairaut's Theorem under conditions of continuity for the second derivatives, reflecting the consistent nature of multivariable calculus.

Differentiation

Differentiation, one of the fundamental operations in calculus, enables us to determine the rate at which one quantity changes in relation to another. With partial differentiation, we apply this operation to functions of multiple variables, treating all but one variable as constants.
In this exercise, partial differentiation allowed us to tackle a function \( f(x, y) \) where each variable can change independently. By differentiating with respect to a particular variable while keeping others constant, we can:

• Understand how changes in one variable directly affect the function output, holding other factors steady.
• Calculate first partial derivatives to find immediate, linear rates of change.
• Proceed to second-order partial derivatives for deeper insight into function behavior.

This method provides a robust framework to dissect complex problems, simplifying the analysis of real-world phenomena where multiple inputs influence outcomes in intricate, interrelated ways.