Third Derivative of Inverse Function
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Theorem
Let $f$ be a real function which is of differentiability class $3$.
Let $f$ have an inverse $f^{-1}$, likewise of differentiability class $3$.
Then:
- $\dfrac {\d^3 x} {\d y^3} = -\paren {\dfrac {\d^3 y} {\d x^3} \dfrac {\d y} {\d x} - 3 \paren {\dfrac {\d^2 y} {\d x^2} }^2} \paren {\dfrac {\d y} {\d x} }^{-5}$
Proof
\(\ds \dfrac {\d^3 x} {\d y^3}\) | \(=\) | \(\ds \map {\dfrac \d \d y } {\dfrac {\d^2 x} {\d y^2} }\) | Definition of Third Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\dfrac \d {\d y} } {-\dfrac {\d^2 y} {\d x^2} \paren {\dfrac {\d y} {\d x} }^{-3} }\) | Derivative of Inverse Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {\dfrac {\d y} {\d x} }^{-3} \map {\dfrac \d {\d y} } {\dfrac {\d^2 y} {\d x^2} } + \dfrac {\d^2 y} {\d x^2} \map {\dfrac \d {\d y} } {\paren {\dfrac {\d y} {\d x} }^{-3} } }\) | Product Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {\dfrac {\d y} {\d x} }^{-3} \dfrac {\d x} {\d y} \map {\dfrac \d {\d x} } {\dfrac {\d^2 y} {\d x^2} } + \dfrac {\d^2 y} {\d x^2} \dfrac {\d x} {\d y} \map {\dfrac \d {\d x} } {\paren {\dfrac {\d y} {\d x} }^{-3} } }\) | Derivative of Composite Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {\dfrac {\d y} {\d x} }^{-3} \dfrac {\d x} {\d y} \dfrac {\d^3 y} {\d x^3} + \dfrac {\d^2 y} {\d x^2} \dfrac {\d x} {\d y} \paren {-3 \paren {\dfrac {\d y} {\d x} }^{-4} } \map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} } }\) | Definition of Third Derivative, Derivative of Composite Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\paren {\dfrac {\d y} {\d x} }^{-3} \dfrac {\d x} {\d y} \dfrac {\d^3 y} {\d x^3} + \paren {\dfrac {\d^2 y} {\d x^2} }^2 \dfrac {\d x} {\d y} \paren {-3 \paren {\dfrac {\d y} {\d x} }^{-4} } }\) | Definition of Second Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\dfrac {\d^3 y} {\d x^3} - 3 \paren {\dfrac {\d^2 y} {\d x^2} }^2 \paren {\dfrac {\d y} {\d x} }^{-1} } \paren {\dfrac {\d y} {\d x} }^{-3} \dfrac {\d x} {\d y}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\dfrac {\d^3 y} {\d x^3} - 3 \paren {\dfrac {\d^2 y} {\d x^2} }^2 \paren {\dfrac {\d y} {\d x} }^{-1} } \paren {\dfrac {\d y} {\d x} }^{-3} \paren {\dfrac {\d y} {\d x} }^{-1}\) | Derivative of Composite Function | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {\dfrac {\d^3 y} {\d x^3} \dfrac {\d y} {\d x} - 3 \paren {\dfrac {\d^2 y} {\d x^2} }^2} \paren {\dfrac {\d y} {\d x} }^{-5}\) | simplifying and manipulating into required form |
$\blacksquare$
Sources
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Leibniz's Theorem for Differentiation of a Product: $3.3.11$