こんにちは、リンス(@Lins016)です。
今回は積と商の導関数（微分）の定義の証明について学習していこう。
スポンサーリンク
積と商の導関数（微分）
積と商の導関数（微分）
・微分の定義
\(\small{ \ f'(x)=\displaystyle\lim_{h \to 0}\displaystyle\frac{f(x+h)-f(x)}{h} \ }\)
・積の微分
\(\small{ \ \left\{f(x)g(x)\right\}’=f'(x)g(x)+f(x)g'(x) \ }\)
\(\small{ \ \left\{f(x)g(x)h(x)\right\}’=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x) \ }\)
・商の微分
\(\small{ \ \left\{\displaystyle \frac{f(x)}{g(x)}\right\}' =\displaystyle \frac{f'(x)g(x)-f(x)g'(x)}{\left\{g(x)\right\}^2} \ }\)
\(\small{ \ \left\{\displaystyle \frac{1}{g(x)}\right\}' =-\displaystyle \frac{g'(x)}{\left\{g(x)\right\}^2} \ }\)
導関数の定義
\(\small{ \ f'(x)=\displaystyle\lim_{h \to 0}\displaystyle\frac{f(x+h)-f(x)}{h} \ }\)
▼あわせてCHECK▼(別ウィンドウで開きます)
二つの関数の積の微分
\(\small{ \ \left\{f(x)g(x)\right\}’\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}\\[20pt]
=\displaystyle\lim_{h \to 0}\left\{\displaystyle\frac{f(x+h)-f(x)}{h}\cdot g(x+h)+ f(x) \cdot\displaystyle\frac{g(x+h)-g(x)}{h}\right\}\\[20pt]
=f'(x)g(x)+f(x)g'(x) \ }\)
三つの関数の積の微分
\(\small{ \ \left\{f(x)g(x)h(x)\right\}’\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle\frac{f(x+h)g(x+h)h(x+h)-f(x)g(x)h(x)}{h}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle\frac{f(x+h)g(x+h)h(x+h)-f(x)g(x+h)h(x+h)+f(x)g(x+h)h(x+h)-f(x)g(x)h(x)}{h}\\[20pt]
=\displaystyle\lim_{h \to 0}\left\{\displaystyle\frac{f(x+h)-f(x)}{h}\cdot g(x+h)h(x+h)+ f(x) \cdot\displaystyle\frac{g(x+h)h(x+h)-g(x)h(x)}{h}\right\}\\[20pt]
=\displaystyle\lim_{h \to 0}\left\{\displaystyle\frac{f(x+h)-f(x)}{h}\cdot g(x)h(x)+ f(x) \cdot\displaystyle\frac{g(x+h)h(x+h)-g(x)h(x+h)+g(x)h(x+h)-g(x)h(x)}{h}\right\}\\[20pt]
=\displaystyle\lim_{h \to 0}\left\{\displaystyle\frac{f(x+h)-f(x)}{h}\cdot g(x+h)h(x+h)+ f(x) \cdot\displaystyle\frac{g(x+h)-g(x)}{h}\cdot h(x+h)+f(x+h) \cdot g(x) \cdot \displaystyle\frac{h(x+h)-h(x)}{h}\right\}\\[20pt]
=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x) \ }\)
商の微分
\(\small{ \ \left\{\displaystyle \frac{f(x)}{g(x)}\right\}' \\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{\displaystyle \frac{f(x+h)}{g(x+h)}-\displaystyle \frac{f(x)}{g(x)}}{h}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{1}{h}\left\{ \displaystyle \frac{f(x+h)}{g(x+h)}-\displaystyle \frac{f(x)}{g(x)}\right\}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{1}{h}\cdot \displaystyle \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{1}{g(x+h)g(x)}\cdot \displaystyle \frac{f(x+h)g(x)-f(x)g(x)-\left\{f(x)g(x+h)-f(x)g(x)\right\}}{h}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{1}{g(x+h)g(x)}\left\{ \displaystyle \frac{f(x+h)-f(x)}{h}\cdot g(x)-f(x)\cdot\displaystyle \frac{g(x+h)-g(x)}{h}\right\}\\[20pt]
=\displaystyle \frac{f'(x)g(x)-f(x)g'(x)}{\left\{g(x)\right\}^2}
\ }\)
\ }\)
分子が定数の商の微分
\(\small{ \ \left\{\displaystyle \frac{1}{g(x)}\right\}' \\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{\displaystyle \frac{1}{g(x+h)}-\displaystyle \frac{1}{g(x)}}{h}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{1}{h}\left\{ \displaystyle \frac{1}{g(x+h)}-\displaystyle \frac{1}{g(x)}\right\}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{1}{h}\cdot \displaystyle \frac{g(x)-g(x+h)}{g(x+h)g(x)}\\[20pt]
=\displaystyle\lim_{h \to 0}\displaystyle \frac{1}{g(x+h)g(x)}\cdot \displaystyle \frac{g(x)-g(x+h)}{h}\\[20pt]
=\displaystyle\lim_{h \to 0}-\displaystyle \frac{1}{g(x+h)g(x)}\cdot\displaystyle \frac{g(x+h)-g(x)}{h}\\[20pt]
=-\displaystyle \frac{g'(x)}{\left\{g(x)\right\}^2}
\ }\)
\ }\)
