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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $f \\colon U \\to \\mathbb{R}$, and let $a \\in U$ be a [[local extremum|04CPG9K1C4]] of $f$.\nIf the [[partial derivative|05AXG3QRKR]] $\\partial f / \\partial x_i (a)$ exists for some $i$, then $\\partial f / \\partial x_i (a) = 0$.\nIn particular, if $f$ is [[differentiable|0135XZZQ8M]] at $a$, then $\\nabla f(a) = 0$ and the derivative $Df(a)$ is the zero linear map.\n\n## Remarks\n\nA point $a \\in U$ where $f$ is differentiable and $\\nabla f(a) = 0$ is a critical point, or stationary point, of $f$.\nThe proof applies the one-variable theorem to $t \\mapsto f(a + t e_i)$, where $e_i$ is the $i$-th standard basis vector. This function has a local extremum at $t = 0$.\nThe converse is false. For $f(x, y) = x^2 - y^2$, $\\nabla f(0, 0) = 0$, but $(0, 0)$ is a [[saddle point|00N3T18QNA]], not an extremum.\nThe hypothesis that $U$ is open matters. On $[0, 1]^2$, $f(x, y) = x + y$ has a maximum at $(1, 1)$, where $\\nabla f = (1, 1)$.\n\n## Explanations\n\n### L1 human\n\nAt the top of a smooth hill the ground is flat in every direction.\nSo all partial derivatives are zero there, and the tangent plane to the graph is horizontal.\nThis turns the search for extrema into the solution of the equations $\\nabla f = 0$.\n",
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