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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $f \\colon U \\to \\mathbb{R}^m$, and let $a \\in U$.\nSuppose that $f$ is [[differentiable|0135XZZQ8M]] at $a$ with derivative $Df(a)$.\nThen for each $v \\in \\mathbb{R}^n$, $f$ [[has the directional derivative|05EGFBZTGC]] $Df(a)(v)$ at $a$ in the direction $v$:\n$$\\lim_{t \\to 0} \\frac{f(a + t v) - f(a)}{t} = Df(a)(v).$$\n\n## Remarks\n\nFor $v = e_i$ this gives $\\dfrac{\\partial f}{\\partial x_i}(a) = Df(a)(e_i)$. So the [[partial derivatives|09EGC6F8PT]] of the components of $f$ are the entries of the matrix of $Df(a)$.\nFor a real-valued $f$, it gives $D_v f(a) = \\nabla f(a) \\cdot v$, where $\\nabla f(a)$ is the [[gradient|0ABK9V9Y64]].\nIn particular, $D_v f(a)$ is linear in $v$. A function whose directional derivatives are not linear in $v$ is not differentiable.\n\n## Explanations\n\n### L1 human\n\nOnce you know the linear map $Df(a)$, you know the slope of $f$ in every direction: feed the direction into the map.\nFor $f(x, y) = x^2 + y^2$ at $(1, 1)$, $Df(1, 1)(v) = 2 v_1 + 2 v_2$. So the slope in the direction $(1, -1)$ is $0$.\n",
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        "locator": "For differentiable functions",
        "url": "https://en.wikipedia.org/wiki/Directional_derivative"
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        "locator": "Finite dimensions",
        "url": "https://en.wikipedia.org/wiki/Fr%C3%A9chet_derivative"
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  "name": "Derivative gives the directional derivatives",
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