Math Problem Statement
Show that f_xy(0, 0) ≠ f_yx(0, 0) for the function defined as f(x, y) = { xy if |x| ≥ |y|, -xy if |x| < |y| }.
Solution
The problem asks to show that the partial derivatives and are not equal for the given function, defined piecewise as:
\begin{cases} xy & \text{if } |x| \geq |y| \\ -xy & \text{if } |x| < |y| \end{cases}$$ Let's proceed step by step to compute the mixed partial derivatives $$f_{xy}(0, 0)$$ and $$f_{yx}(0, 0)$$ and show that they are not equal. ### Step 1: Compute the Partial Derivatives We begin by calculating the first-order partial derivatives of the function with respect to $$x$$ and $$y$$, and then the second-order mixed partial derivatives at the point $$(0, 0)$$. #### Case 1: $$|x| \geq |y|$$ Here, the function is $$f(x, y) = xy$$. 1. The partial derivative with respect to $$x$$: $$f_x(x, y) = y$$ 2. The partial derivative with respect to $$y$$: $$f_y(x, y) = x$$ #### Case 2: $$|x| < |y|$$ Here, the function is $$f(x, y) = -xy$$. 1. The partial derivative with respect to $$x$$: $$f_x(x, y) = -y$$ 2. The partial derivative with respect to $$y$$: $$f_y(x, y) = -x$$ ### Step 2: Compute Mixed Partial Derivatives Now, we compute the second-order mixed partial derivatives. #### For $$f_{xy}(0, 0)$$: To compute $$f_{xy}(0, 0)$$, we take the derivative of $$f_x(x, y)$$ with respect to $$y$$. 1. For $$|x| \geq |y|$$, $$f_x(x, y) = y$$, so: $$f_{xy}(x, y) = \frac{\partial}{\partial y} (y) = 1$$ 2. For $$|x| < |y|$$, $$f_x(x, y) = -y$$, so: $$f_{xy}(x, y) = \frac{\partial}{\partial y} (-y) = -1$$ At $$(0, 0)$$, both cases overlap, but the limit behavior in neighborhoods where $$|x| \geq |y|$$ and $$|x| < |y|$$ shows that the value of $$f_{xy}(0, 0)$$ is dependent on the direction of approach. Therefore, the behavior is non-continuous, making $$f_{xy}(0, 0)$$ undefined or non-equal in each region. #### For $$f_{yx}(0, 0)$$: Similarly, to compute $$f_{yx}(0, 0)$$, we take the derivative of $$f_y(x, y)$$ with respect to $$x$$. 1. For $$|x| \geq |y|$$, $$f_y(x, y) = x$$, so: $$f_{yx}(x, y) = \frac{\partial}{\partial x} (x) = 1$$ 2. For $$|x| < |y|$$, $$f_y(x, y) = -x$$, so: $$f_{yx}(x, y) = \frac{\partial}{\partial x} (-x) = -1$$ Again, at $$(0, 0)$$, the value of $$f_{yx}(0, 0)$$ is dependent on the region of approach, and we observe non-continuous behavior, similar to the case of $$f_{xy}$$. ### Conclusion: The key observation is that the mixed partial derivatives $$f_{xy}(0, 0)$$ and $$f_{yx}(0, 0)$$ depend on the direction of approach due to the piecewise nature of the function. Therefore, they are not equal at $$(0, 0)$$. Would you like further details or have any questions on this? Below are some related questions that might help expand your understanding: 1. What are the formal conditions for equality of mixed partial derivatives? 2. How does the definition of differentiability relate to mixed partial derivatives? 3. Can we find other functions where $$f_{xy} \neq f_{yx}$$? 4. What is the geometric interpretation of mixed partial derivatives in 3D space? 5. How would this analysis change if the function were defined differently around $$(0, 0)$$? **Tip:** Always check the continuity of partial derivatives when dealing with piecewise functions to understand their behavior at boundary points.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Partial Derivatives
Multivariable Calculus
Piecewise Functions
Formulas
f_xy(0, 0) and f_yx(0, 0) represent mixed partial derivatives at the origin
Theorems
Clairaut's Theorem on Equality of Mixed Partial Derivatives
Suitable Grade Level
College Level
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