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  "body": "\n## Counterexample\n\nOne could guess that the mixed [[partial derivatives|02RSXVM1ZN]] $f_{xy}$ and $f_{yx}$ are equal at each point where both exist.\nThis is false.\nLet $f \\colon \\mathbb{R}^2 \\to \\mathbb{R}$ be given by\n$$f(x, y) = \\frac{x y (x^2 - y^2)}{x^2 + y^2} \\text{ for } (x, y) \\ne (0, 0), \\qquad f(0, 0) = 0.$$\nBoth mixed second partial derivatives of $f$ exist at $(0, 0)$, but $\\dfrac{\\partial}{\\partial y}\\left(\\dfrac{\\partial f}{\\partial x}\\right)(0, 0) = -1$ and $\\dfrac{\\partial}{\\partial x}\\left(\\dfrac{\\partial f}{\\partial y}\\right)(0, 0) = 1$.\n\n## Remarks\n\nFrom the definition, $\\partial f / \\partial x (0, y) = \\lim_{h \\to 0} f(h, y) / h = -y$ for each $y$, and $\\partial f / \\partial y (x, 0) = \\lim_{k \\to 0} f(x, k) / k = x$ for each $x$. Differentiate these once more at $0$.\nThe function is of class $C^1$ on $\\mathbb{R}^2$, but its second partial derivatives are not continuous at the origin. So [[Schwarz's theorem|0NNQY8WCW6]] does not apply.\nThis example is due to Peano.\n\n## Explanations\n\n### L1 human\n\nSchwarz's theorem needs continuous second partial derivatives, and this function shows why.\nAway from the origin everything is smooth, but at the origin the two orders of differentiation see different limits.\n",
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