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関数の積の微分

${\displaystyle {\left\{g\left(x\right)h\left(x\right)\right\}}^{\prime }}$${\displaystyle =g^{\prime }\left(x\right)h\left(x\right)+g\left(x\right)h^{\prime }\left(x\right)}$

すなわち，

${\displaystyle f\left(x\right)=g\left(x\right)h\left(x\right)}$${\displaystyle \to f^{\prime }\left(x\right)=g^{\prime }\left(x\right)h\left(x\right)+g\left(x\right)h{\left(x\right)}^{\prime }}$

■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{1}{h}\left\{g\left(x+h\right)h\left(x+h\right)\right.}$${\displaystyle \left.-g\left(x\right)h\left(x\right)\right\}}$

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

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

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

${\displaystyle =g^{\prime }\left(x\right)h\left(x\right)+g\left(x\right)h^{\prime }\left(x\right)}$

よって，

${\displaystyle {\left\{g\left(x\right)h\left(x\right)\right\}}^{\prime }}$${\displaystyle =g^{\prime }\left(x\right)h\left(x\right)+g\left(x\right)h^{\prime }\left(x\right)}$

である．

 

ホーム>>カテゴリー分類>>微分>>導関数の基本式I関数の積の微分

最終更新日： 2018年3月31日

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