Second Derivative of Inverse Function

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Theorem

Let $f$ be a real function which is of differentiability class $2$.

Let $f$ have an inverse $f^{-1}$, likewise of differentiability class $2$.

Then:

$\dfrac {\d^2 x} {\d y^2} = -\dfrac {\d^2 y} {\d x^2} \paren {\dfrac {\d y} {\d x} }^{-3}$


Proof

\(\ds \dfrac {\d^2 x} {\d y^2}\) \(=\) \(\ds \map {\dfrac \d {\d y} } {\dfrac {\d x} {\d y} }\) Definition of Second Derivative
\(\ds \) \(=\) \(\ds \map {\dfrac \d {\d y} } {\dfrac 1 {\d y / \d x} }\) Derivative of Inverse Function
\(\ds \) \(=\) \(\ds \dfrac {-1} {\paren {\d y / \d x}^2} \map {\dfrac \d {\d y} } {\dfrac {\d y} {\d x} }\) Derivative of Composite Function
\(\ds \) \(=\) \(\ds \dfrac {-1} {\paren {\d y / \d x}^2} \dfrac {\d x} {\d y} \map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} }\) Derivative of Composite Function
\(\ds \) \(=\) \(\ds \dfrac {-1} {\paren {\d y / \d x}^3} \dfrac {\d^2 y} {\d x^2}\) Definition of Second Derivative, Derivative of Inverse Function and simplification
\(\ds \) \(=\) \(\ds -\dfrac {\d^2 y} {\d x^2} \paren {\dfrac {\d y} {\d x} }^{-3}\) rearranging into given format

$\blacksquare$


Sources

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