逆函数の微分

微分商形式

$ {f^{-1}}'(x)=\frac{1}{f'(f^{-1}(x))}

微分形式

$ \mathrm df^{-1}(x)=\frac{1}{f'(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