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  "body": "\n## Counterexample\n\nOne could guess that a function of several variables is differentiable at a point when all its partial derivatives exist there.\nThis is false.\nLet $f \\colon \\mathbb{R}^2 \\to \\mathbb{R}$ be given by $f(x, y) = \\dfrac{xy}{x^2 + y^2}$ for $(x, y) \\ne (0, 0)$ and $f(0, 0) = 0$.\nBoth [[partial derivatives|09EGC6F8PT]] of $f$ exist at $(0, 0)$, and both are $0$.\nBut $f$ is not continuous at $(0, 0)$, so $f$ is not [[differentiable|0HHHDCXZV4]] at $(0, 0)$.\n\n## Remarks\n\nThe function is $0$ on both axes, so $f(h, 0) = f(0, h) = 0$ for each $h$, and both difference quotients at the origin are $0$.\nOn the line $y = x$ the value is $1/2$, so $f$ is not continuous at the origin, as in the [[separately continuous function|0C0KPAKK9Y]]. A function that is differentiable at a point is continuous there.\nEven all directional derivatives are not enough. For $g(x, y) = x^2 y / (x^4 + y^2)$ off the origin and $g(0, 0) = 0$, each directional derivative at the origin exists, but $g = 1/2$ on the parabola $y = x^2$ minus the origin.\nIf the partial derivatives exist near the point and are continuous at it, then the function is differentiable, by the [[criterion with continuous partial derivatives|0KXSK8TEEB]].\n\n## Explanations\n\n### L1 human\n\nThe partial derivatives look only along the two axes.\nA function can be flat on both axes and still do something wild between them, here on the diagonal.\nDifferentiability asks for a good linear approximation in all directions at once, which is much more.\n",
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