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  "body": "\n## Statement\n\nLet $I \\subseteq \\mathbb{R}$ be an open interval, let $U \\subseteq \\mathbb{R}^n$ be open, let $\\gamma \\colon I \\to U$, let $f \\colon U \\to \\mathbb{R}^m$, and let $t_0 \\in I$.\nSuppose that $\\gamma$ is differentiable at $t_0$ and $f$ is differentiable at $\\gamma(t_0)$.\nThen $f \\circ \\gamma$ is differentiable at $t_0$, and\n$$(f \\circ \\gamma)'(t_0) = Df(\\gamma(t_0))\\big(\\gamma'(t_0)\\big).$$\n\n## Remarks\n\nFor $m = 1$, this reads $(f \\circ \\gamma)'(t_0) = \\nabla f(\\gamma(t_0)) \\cdot \\gamma'(t_0) = \\sum_{i=1}^n \\dfrac{\\partial f}{\\partial x_i}(\\gamma(t_0)) \\, \\gamma_i'(t_0)$, with the [[gradient|0ABK9V9Y64]] $\\nabla f$.\nIt is the case $n = 1$ of the [[chain rule in several variables|0EAAS2EHDR]], since a linear map $\\mathbb{R} \\to \\mathbb{R}^n$ is $s \\mapsto s \\gamma'(t_0)$.\nFor $\\gamma(t) = a + t v$ it gives the [[directional derivative|06KWGHDN2J]] $D_v f(a) = Df(a)(v)$.\n\n## Explanations\n\n### L1 human\n\nWalk along a [[curve|0G7WM8DX4P]] $\\gamma(t)$ through a landscape with height $f$.\nThe rate at which your height changes is the slope of the landscape in the direction you walk, times your speed.\nFor $f(x, y) = x y$ and $\\gamma(t) = (t, t^2)$, the rule gives $\\nabla f \\cdot \\gamma' = t^2 \\cdot 1 + t \\cdot 2t = 3 t^2$, the derivative of $t^3$.\n",
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