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6 対数関数の微分

定理 3.20 (対数関数の微分)  

$\displaystyle \frac{d}{dx}\,\log x=\frac{1}{x}$ (387)

問 3.21   これを示せ.


(証明) $ y=f(x)=\log x$ とおき定義に従い計算すると,

$\displaystyle \frac{dy}{dx}$ $\displaystyle = f'(x)= \lim_{h\to0}\frac{f(x+h)-f(x)}{h}= \lim_{h\to0}\frac{\log(x+h)-\log(x)}{h}$ (388)
  $\displaystyle = \lim_{h\to0}\frac{1}{h}\left(\log(x+h)-\log(x)\right) = \lim_{h...
...{h}\log\frac{x+h}{x} = \lim_{h\to0}\log\left(\frac{x+h}{x}\right)^{\frac{1}{h}}$ (389)
  $\displaystyle = \lim_{h\to0}\log\left(1+\frac{h}{x}\right)^{\frac{1}{h}} = \lim...
...0} \frac{1}{x}\log \left(1+\frac{1}{\quad\frac{x}{h}\quad}\right)^{\frac{x}{h}}$ (390)
  $\displaystyle = \frac{1}{x}\log\left\{ \lim_{h\to0} \left(1+\frac{1}{\quad\frac...
...og\left\{ \lim_{z=\frac{x}{h}\to\infty} \left(1+\frac{1}{z}\right)^{z} \right\}$ (391)
  $\displaystyle = \frac{1}{x}\log e = \frac{1}{x}$ (392)

を得る.



Kondo Koichi
Created at 2003/08/29