分数関数の微分 I

${\displaystyle {\left\{\frac{1}{g\left(x\right)}\right\}}^{\prime }=-\frac{g^{\prime }\left(x\right)}{{\left\{g\left(x\right)\right\}}^2}}$

すなわち

${\displaystyle f\left(x\right)=\frac{1}{g\left(x\right)}\to f^{\prime }\left(x\right)=-\frac{g^{\prime }\left(x\right)}{{\left\{g\left(x\right)\right\}}^2}}$

■導出

${\displaystyle f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}}$

${\displaystyle =\underset{h\to 0}{lim}\frac{\frac{1}{g\left(x+h\right)}-\frac{1}{g\left(x\right)}}{h}}$

${\displaystyle =\underset{h\to 0}{lim}\frac{\frac{g\left(x\right)-g\left(x+h\right)}{g\left(x+h\right)g\left(x\right)}}{h}}$

${\displaystyle =\underset{h\to 0}{lim}\left\{\frac{-1}{g\left(x+h\right)g\left(x\right)}\right.}$${\displaystyle \left.·\frac{g\left(x+h\right)-g\left(x\right)}{h}\right\}}$

${\displaystyle =\left\{\underset{h\to 0}{lim}\frac{-1}{g\left(x+h\right)g\left(x\right)}\right\}}$${\displaystyle \left\{\underset{h\to 0}{lim}\frac{g\left(x+h\right)-g\left(x\right)}{h}\right\}}$

${\displaystyle =-\frac{g^{\prime }\left(x\right)}{{\left\{g\left(x\right)\right\}}^2}}$    ここを参照

よって

${\displaystyle {\left\{\frac{1}{g\left(x\right)}\right\}}^{\prime }=-\frac{g^{\prime }\left(x\right)}{{\left\{g\left(x\right)\right\}}^2}}$

である．

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最終更新日 2023年6月7日