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  "body": "\n## Definition\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $f \\colon U \\to \\mathbb{R}$, and let $a \\in U$ be a point at which $f$ is [[differentiable|0135XZZQ8M]].\nThe gradient of $f$ at $a$ is the unique vector $\\nabla f(a) \\in \\mathbb{R}^n$ such that\n$$Df(a)(v) = \\nabla f(a) \\cdot v \\quad \\text{for each } v \\in \\mathbb{R}^n,$$\nwhere $\\cdot$ is the dot product of [[Euclidean space|0XPR1E85W0]].\nIn coordinates, it is the vector of [[partial derivatives|09EGC6F8PT]]\n$$\\nabla f(a) = \\left(\\frac{\\partial f}{\\partial x_1}(a), \\dots, \\frac{\\partial f}{\\partial x_n}(a)\\right).$$\nOther notations are $\\operatorname{grad} f(a)$ and $\\nabla f(a)$, where $\\nabla$ is the [[del operator|0H6DP8CHND]].\n\n## Remarks\n\nThe vector exists and is unique because each linear map $\\mathbb{R}^n \\to \\mathbb{R}$ is $v \\mapsto w \\cdot v$ for exactly one $w \\in \\mathbb{R}^n$.\nFor $f(x, y) = x^2 + 3 x y$, the gradient is $\\nabla f(x, y) = (2x + 3y, \\, 3x)$.\nA gradient is a column vector, and the derivative $Df(a)$ is the row vector of the same entries. Some books do not make this distinction.\nThe map $a \\mapsto \\nabla f(a)$ is a [[vector field|0299VCW1TH]] on $U$, the gradient field of $f$.\n\n## Explanations\n\n### L1 human\n\nThe gradient collects all slopes of $f$ along the axes into one arrow.\nOn a map of a hill, the gradient at a point is an arrow that points straight uphill. The arrow is longer where the hill is steeper.\n",
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