Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x, y)=x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}}\) if \((x, y) \neq(0,0) \quad\) and \(f(0,0)=0 .\) Show that \(f_{x y}(0,0) \neq f_{y x}(0,0)\) by completing the following steps: (a) Show that \(f_{x}(0, y)=\lim _{h \rightarrow 0} \frac{f(0+h, y)-f(0, y)}{h}=-y\) for all \(y\). (b) Similarly, show that \(f_{y}(x, 0)=x\) for all \(x\). (c) Show that \(f_{y x}(0,0)=\lim _{h \rightarrow 0} \frac{f_{y}(0+h, 0)-f_{y}(0,0)}{h}=1\). (d) Similarly, show that \(f_{x y}(0,0)=-1\).

Step-by-Step Solution

Answer

Yes, \(f_{xy}(0,0) = -1\) and \(f_{yx}(0,0) = 1\), therefore they are not equal.

Step 1: Show that \(f_x(0, y) = -y\)

To find \(f_x(0, y)\), we evaluate the limit \( \lim_{h \to 0} \frac{f(h, y) - f(0, y)}{h} \). Since \(f(0, y) = 0\),\[f(h, y) = hy \frac{h^2 - y^2}{h^2 + y^2}\]Simplifying the above expression,\[\frac{h^2 - y^2}{h^2 + y^2} = \frac{h^2}{h^2 + y^2} - \frac{y^2}{h^2 + y^2}\]Consider the first term:\[\lim_{h \to 0} hy \frac{h^2}{h^2 + y^2} = \lim_{h \to 0} hy \left(1 - \frac{y^2}{h^2 + y^2}\right) = hy\]Since \(h \to 0\), this term tends to zero. For the second term,\[\lim_{h \to 0} hy \left(-\frac{y^2}{h^2 + y^2}\right) = -y\]Therefore, \(f_x(0, y) = -y\).

Step 2: Show that \(f_y(x, 0) = x\)

To find \(f_y(x, 0)\), evaluate the limit \( \lim_{h \to 0} \frac{f(x, h) - f(x, 0)}{h} \). Since \(f(x, 0) = 0\),\[f(x, h) = xh \frac{x^2 - h^2}{x^2 + h^2} \approx x\]For the expression,\[\frac{x^2 - h^2}{x^2 + h^2} = \frac{x^2}{x^2 + h^2} - \frac{h^2}{x^2 + h^2}\]The first term simplifies to:\[\lim_{h \to 0} xh \left(\frac{x^2}{x^2 + h^2}\right) = x\]The second term contributes insignificantly as \(h \to 0\). Thus, \(f_y(x, 0) = x\).

Step 3: Show that \(f_{y x}(0,0) = 1\)

Evaluate the limit \( \lim_{h \to 0} \frac{f_y(h, 0) - f_y(0, 0)}{h} \). We already found \(f_y(x, 0) = x\). So,\[f_y(h, 0) = h \quad ext{and} \quad f_y(0, 0) = 0\]Thus,\[f_{y x}(0, 0) = \lim_{h \to 0} \frac{h - 0}{h} = 1\]

Step 4: Show that \(f_{x y}(0,0) = -1\)

Evaluate the limit \( \lim_{h \to 0} \frac{f_x(0, h) - f_x(0, 0)}{h} \). We determined that \(f_x(0, y) = -y\). Therefore,\[f_x(0, h) = -h \quad ext{and} \quad f_x(0, 0) = 0\]So,\[f_{x y}(0, 0) = \lim_{h \to 0} \frac{-h - 0}{h} = -1\]

Key Concepts

Partial Derivatives

Partial derivatives are essential when dealing with functions of multiple variables. Imagine a scenario where you have a surface represented by the function \( f(x, y) \). To understand how this surface changes as you adjust each variable separately, you use partial derivatives.
To take a partial derivative with respect to \( x \), you treat \( y \) as a constant, and vice versa when different distances in the \( x \)-dependence and \( y \)-dependence. This way, you analyze how \( f(x, y) \) changes when moving along the respective axis in a given plane.

• First Partial Derivative with Respect to x: Differentiate \( f(x, y) \) with respect to \( x \), holding \( y \) constant.
• First Partial Derivative with Respect to y: Similar approach; differentiate \( f(x, y) \) with respect to \( y \), keeping \( x \) constant.

In higher-level calculus, such as in this problem, you encounter second partial derivatives. These explore the rate of change in the rate of change, providing deeper insights into the function's behavior.
This problem reveals unique nuances in mixed partials \( f_{xy} \) and \( f_{yx} \), showcasing that \( f_{xy}(0,0) \) is not necessarily equal to \( f_{yx}(0,0) \). This deviation happens due to the specific nature of the function at points such as \( (0, 0) \).

Limits

When exploring calculus, you frequently encounter limits, especially while working with functions and derivatives. Limits help you understand the behavior of a function near a specific point. In the context of partial derivatives, limits are crucial in determining the derivative's existence and value.

• Concept of Limits: A limit represents what a function approaches as its input gets closer to a given value.
• Limits in Derivatives: In finding derivatives, the idea is to measure the slope of the tangent line to the curve. This requires evaluating limits as the difference in the function values approaches zero.

In the provided example, you calculate these limits to see the unique nature of mixed partial derivatives. Calculating \( \lim_{h \to 0} \frac{f(h, y) - f(0, y)}{h} \) describes the rate of change concerning \( x \), and similarly for \( y \). While limits help find these derivatives, they are equally important in understanding the continuity and precision of multi-variable functions.
These calculations show how, although a function seems continuous, its derivatives may not behave as intuitively expected, especially at defined crucial points like \( (0,0) \).

Functions of Several Variables

In calculus, analyzing functions of several variables is imperative. Such functions help define and model surfaces or spaces in more than just one dimension. They allow the exploration of equations like \( z = f(x, y) \), providing insights beyond the realm of a single-line graph.
When dealing with multiple variables:

• Complexity in Relationships: Each variable brings added complexity. How one variable impacts the function can intricately depend on another.
• Multi-Dimensional Analysis: These functions are not fixed on a single plane – they involve spatial changes across multiple axes, akin to examining hills and valleys in a landscape.
• Interactions: Partial derivatives highlight the unique interactions between these variables. They assist in understanding how infinitesimally small tweaks in one variable affect the whole function.

Analyzing \( f(x, y) = xy \frac{x^2 - y^2}{x^2 + y^2} \) unveils significant interactions at the core, notably at the origin where behaviors are not symmetrical. Derivative mismatches like \( f_{xy}(0,0) eq f_{yx}(0,0) \) arise from these intricate dependencies, showcasing that the influence exerted by \( x \) and \( y \) can vary based on their interdependencies within the function.