逆函数の微分
微分商形式
{f^{-1}}'(x)=\frac{1}{f'\circ f^{-1}(x)}
微分形式
\mathrm df^{-1}(x)=\frac{1}{f'\circ f^{-1}(x)}\mathrm dx
多変数函数
\bm{\nabla}\bm{f}^{-1}=\left(\left.\bm{\nabla}\bm{f}(\bm{x})\right|_{\bm{x}=\bm{f}^{-1}}\right)^{-1}
導出
x=f\circ f^{-1}(x)
\implies \mathrm dx=\mathrm df\circ f^{-1}(x)
= f'\circ f^{-1}(x)\mathrm df^{-1}(x)
\underline{\implies\mathrm df^{-1}(x)=\frac{1}{f'\circ f^{-1}(x)}\mathrm dx\quad}_\blacksquare
多変数函数版の導出
\bm{f}\circ\bm{f}^{-1}(\bm{x})=\bm{x}
\implies \bm{I}=\bm{\nabla}(\bm{f}\circ\bm{f}^{-1}(\bm{x}))=\left.\bm{\nabla}\bm{f}(\bm{y})\right|_{\bm{y}=\bm{f}^{-1}(\bm{x})}\bm{\nabla}\bm{f}^{-1}(\bm{x})
\underline{\implies \bm{\nabla}\bm{f}^{-1}=\left(\left.\bm{\nabla}\bm{f}(\bm{x})\right|_{\bm{x}=\bm{f}^{-1}}\right)^{-1}\quad}_\blacksquare