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# Symmetry of Mixed Partial Derivatives (Clairaut’s Theorem)
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:::{prf:theorem} Clairaut Schwarz
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:label: thm-Clairaut-appendix
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:nonumber:
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Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a function such that both mixed partial derivatives $\frac{\partial^2 f}{\partial x \partial y}$ and $\frac{\partial^2 f}{\partial y \partial x}$ exist and are **continuous** on an open set containing a point $(x_0, y_0)$
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Then:
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$$
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\boxed{
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\frac{\partial^2 f}{\partial x \partial y}(x_0, y_0) = \frac{\partial^2 f}{\partial y \partial x}(x_0, y_0)
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}
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$$
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That is, **the order of differentiation can be interchanged**.
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:::
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## Intuition
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If a function is smooth enough (specifically, if the second-order partial derivatives exist and are continuous), then the "curvature" in the $x$ direction after differentiating in the $y$ direction is the same as the curvature in the $y$ direction after differentiating in the $x$ direction.
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---
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## Proof Sketch
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We will sketch a proof using the **mean value theorem** and the definition of partial derivatives. Let’s assume that $f$ has continuous second partial derivatives in an open rectangle around the point $(x_0, y_0)$.
Then, as $h, k \to 0$, $F(h,k) \to \frac{\partial^2 f}{\partial y \partial x}(x_0, y_0)$ and also $F(h,k) \to \frac{\partial^2 f}{\partial x \partial y}(x_0, y_0)$, provided the second partial derivatives are continuous.
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### Step-by-step:
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1. By the **Mean Value Theorem**, the numerator of $F(h,k)$ can be interpreted as a finite difference approximation to a mixed partial derivative.
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2. Using Taylor’s Theorem with remainder, or via integral representations of derivatives, one can show that:
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