Problem 30 - Calculus with Concepts in Calculus

Question

Find \(f_{x y}\) and \(f_{y x}\) $$ f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

Both \(f_{xy}\) and \(f_{yx}\) are equal to \(\frac{8x(x^2 - y^2)}{(x^2 + y^2)^3}\).

Step 1: Differentiate with respect to x (First Partial)

To find the mixed partial derivative \(f_{xy}\), we need the first partial derivative of the function \(f(x, y) = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}\) with respect to \(x\). Use the quotient rule for differentiation: if \(v(x, y) = g(x, y)/h(x, y)\), then \(v'_x(x, y) = \frac{g'_x h - g h'_x}{h^2}\). Here, \(g(x, y) = x^2 - y^2\) and \(h(x, y) = x^2 + y^2\). Differentiating them, we get \(g'_x = 2x\) and \(h'_x = 2x\). Substituting: \[ f_{x} = \frac{(2x)(x^2 + y^2) - (x^2 - y^2)(2x)}{(x^2 + y^2)^2} \] Simplifying, we get: \[ f_{x} = \frac{2x(x^2 + y^2 - x^2 + y^2)}{(x^2 + y^2)^2} = \frac{4xy^2}{(x^2 + y^2)^2} \]

Step 2: Differentiate with respect to y (Second Partial of First)

Now differentiate \(f_x = \frac{4xy^2}{(x^2 + y^2)^2}\) with respect to \(y\) to find \(f_{xy}\). Using the product and quotient rules, we have: \[ f_{xy} = \frac{4x(2y(x^2 + y^2)^2 - 4xy^2(x^2 + y^2)2y)}{(x^2 + y^2)^4} \] Simplify this expression to find: \[ f_{xy} = \frac{8x(x^2 - y^2)}{(x^2 + y^2)^3} \]

Step 3: Differentiate with respect to y (First Partial)

To find \(f_{yx}\), first take the partial derivative of \(f(x, y) = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}\) with respect to \(y\). Again, apply the quotient rule: \[ f_{y} = \frac{(-2y)(x^2 + y^2) - (x^2 - y^2)(2y)}{(x^2 + y^2)^2} \] Simplifying gives: \[ f_{y} = \frac{-2y(x^2 + y^2 - x^2 + y^2)}{(x^2 + y^2)^2} = \frac{-4yx^2}{(x^2 + y^2)^2} \]

Step 4: Differentiate with respect to x (Second Partial of Second)

Finally, differentiate \(f_y = \frac{-4yx^2}{(x^2 + y^2)^2}\) with respect to \(x\) to find \(f_{yx}\). This involves the product and quotient rules again: \[ f_{yx} = \frac{(4x(2y(x^2 + y^2)^2 - 4yx^2(x^2 + y^2)2x)}{(x^2 + y^2)^4} \] Simplifying gives: \[ f_{yx} = \frac{8x(x^2 - y^2)}{(x^2 + y^2)^3} \]

Key Concepts

Quotient Rule

The quotient rule is a fundamental technique in calculus for differentiating functions that are expressed as a ratio, or quotient, of two other functions. When you have a function like \( v(x, y) = \frac{g(x, y)}{h(x, y)} \), the quotient rule allows you to find its derivative with respect to a variable, such as \(x\) or \(y\).

The formula for the derivative of the quotient \( v(x, y) \) with respect to \( x \) is given by:

• \( v'_x(x, y) = \frac{g'_x h - g h'_x}{h^2} \)

Here, \( g(x, y) \) and \( h(x, y) \) are the numerator and denominator functions, respectively, and \( g'_x \) and \( h'_x \) are their partial derivatives with respect to \( x \).

The quotient rule is especially crucial when dealing with rational functions, like the one in the exercise \( f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} \). Using this rule helps find the first partial derivatives \( f_x \) and \( f_y \), which are then used in further steps of finding mixed partial derivatives.

Mixed Partial Derivatives

Mixed partial derivatives involve differentiating a function with respect to different variables in sequence. For a function of two variables, like \( f(x, y) \), we can compute derivatives such as \( f_{xy} \) and \( f_{yx} \). These derivatives show how the function changes in both variables and are part of a larger set of second-order derivatives.

In this exercise, the goal is to find \( f_{xy} \) and \( f_{yx} \), which represent the change in the partial derivative of \( f \) with respect to \( x \) when further differentiated with respect to \( y \) and vice versa. Notably, Clairaut's theorem on the equality of mixed partial derivatives tells us that if \( f \) is continuous and its partial derivatives up to a certain order exist, then \( f_{xy} = f_{yx} \). This property is demonstrated in the exercise as both mixed partial derivatives end up giving the same result:

• \( f_{xy} = \frac{8x(x^2 - y^2)}{(x^2 + y^2)^3} \)
• \( f_{yx} = \frac{8x(x^2 - y^2)}{(x^2 + y^2)^3} \)

This symmetry simplifies many problems in calculus, providing a powerful consistency check when calculating derivatives.

Function of Two Variables

Understanding functions of two variables is crucial in multivariable calculus. A function of two variables, say \( f(x, y) \), assigns a real number to each pair of real numbers \( (x, y) \).

These functions create surfaces in a three-dimensional space. For the function in the exercise \( f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} \), the surface it defines can be visualized as showing how the ratio of \( x^2-y^2 \) to \( x^2+y^2 \) changes over the plane.

Partial derivatives, such as \( f_x \) and \( f_y \), measure how the surface's height changes in response to movement along the x-axis or y-axis, respectively. These local changes in direction are foundational concepts that extend into more complex relations and functions in higher dimensions.

• Understanding this helps set the foundation for analyzing and understanding more complex functions of multiple variables.
• It facilitates comprehension of physical phenomena that depend on more than one factor, such as geography, economics, engineering, etc.

Within this framework, calculating derivatives, using techniques like the quotient rule, becomes essential to understanding how variables interact and influence each other.