Chapter 13: Problem 30
Find \(f_{x y}\) and \(f_{y x}\) $$ f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} $$
Short Answer
Expert verified
Both \(f_{xy}\) and \(f_{yx}\) are equal to \(\frac{8x(x^2 - y^2)}{(x^2 + y^2)^3}\).
Step by step solution
01
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: \[ f_{x} = \frac{2x(x^2 + y^2 - x^2 + y^2)}{(x^2 + y^2)^2} = \frac{4xy^2}{(x^2 + y^2)^2} \]
02
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: \[ f_{xy} = \frac{8x(x^2 - y^2)}{(x^2 + y^2)^3} \]
03
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: \[ f_{y} = \frac{-2y(x^2 + y^2 - x^2 + y^2)}{(x^2 + y^2)^2} = \frac{-4yx^2}{(x^2 + y^2)^2} \]
04
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: \[ f_{yx} = \frac{8x(x^2 - y^2)}{(x^2 + y^2)^3} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
The formula for the derivative of the quotient \( v(x, y) \) with respect to \( x \) is given by:
- \( v'_x(x, y) = \frac{g'_x h - g h'_x}{h^2} \)
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:
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:
- \( f_{xy} = \frac{8x(x^2 - y^2)}{(x^2 + y^2)^3} \)
- \( f_{yx} = \frac{8x(x^2 - y^2)}{(x^2 + y^2)^3} \)
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.
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.