{
  "aliases": [
    "inverse-derivative-formula"
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  "body": "\n## Statement\n\nLet $V, W \\subseteq \\mathbb{R}^n$ be open, and let $f \\colon V \\to W$ be a bijection.\nSuppose that $f$ is differentiable at $a \\in V$ and that $f^{-1}$ is differentiable at $b = f(a)$.\nThen $Df(a)$ is invertible, and\n$$D(f^{-1})(b) = \\big( Df(a) \\big)^{-1}.$$\nIn terms of [[Jacobian matrices|0YF0VXKXGB]], $J_{f^{-1}}(b) = J_f(a)^{-1}$.\n\n## Remarks\n\nProof: apply the [[chain rule|0EAAS2EHDR]] to $f^{-1} \\circ f = \\mathrm{id}_V$ and $f \\circ f^{-1} = \\mathrm{id}_W$. This gives $D(f^{-1})(b) \\, Df(a) = I$ and $Df(a) \\, D(f^{-1})(b) = I$.\nThe hard part is the differentiability of $f^{-1}$. The [[inverse function theorem|0K9HSJFCMJ]] gives it when $f$ is of class $C^1$ and $Df(a)$ is invertible.\nFor polar coordinates $f(r, \\theta) = (r \\cos \\theta, r \\sin \\theta)$ with $r \u003e 0$, $J_f = \\begin{pmatrix} \\cos \\theta \u0026 -r \\sin \\theta \\\\ \\sin \\theta \u0026 r \\cos \\theta \\end{pmatrix}$, and the Jacobian matrix of a local inverse is the inverse of this matrix.\n\n## Explanations\n\n### L1 human\n\nUndoing a map undoes its linear approximation.\nSo the derivative of the inverse is the inverse matrix of the derivative, taken at the matching point.\n",
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    "name": "Derivative of the inverse function",
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      {
        "locator": "Several variables",
        "url": "https://en.wikipedia.org/wiki/Inverse_function_theorem"
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  "msc": [
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  "name": "Derivative of the inverse function",
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