分数関数の微分 I

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

すなわち

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

■導出

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

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

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

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

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

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

よって

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

である．

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