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  "body": "\n## Statement\n\nLet $V$ and $W$ be normed vector spaces over $\\mathbb{R}$, let $U \\subseteq V$ be open, let $f \\colon U \\to W$, and let $x \\in U$.\nIf the bounded linear maps $A$ and $B$ from $V$ to $W$ are both [[Fréchet derivatives|08EVNYZF42]] of $f$ at $x$, then $A = B$.\n\n## Remarks\n\nThe proof uses that $U$ is open. For a unit vector $v$ and small $t \u003e 0$, the point $x + t v$ lies in $U$, and $\\|t A v - t B v\\| = o(t)$. Divide by $t$ and let $t \\to 0$ to get $A v = B v$.\nSo it is correct to speak of the derivative $Df(x)$.\nOn a set that is not open, for example a single point, a derivative within the set need not be unique.\n\n## Explanations\n\n### L1 human\n\nTwo different linear maps differ by a definite amount in some direction, at the scale of the step.\nIf both were good approximations, their difference would be tiny compared with the step in every direction. This forces the difference to be $0$.\n",
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