定義に従って，次の関数を微分せよ．
\( (1) \) \( f(x) = x^n \)
\( (2) \) \( f(x) = \sin x \)
\( (3) \) \( f(x) = e^x \)
\( (4) \) \( f(x) = \log x \)
＊導関数＊
関数 \( f(x) \) の導関数は
\( f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \)
\( (1) \)
\( f'(x) = \displaystyle\lim_{h \to 0} \dfrac{(x+h)^n – x^n}{h} \)
\( = \displaystyle\lim_{h \to 0} \dfrac{n x^{n-1} h + {}_n \mathrm{C}_2 x^{n-2} h^2 + \cdots + h^n}{h} \)
\( = \displaystyle\lim_{h \to 0} (n x^{n-1} + {}_n \mathrm{C}_2 x^{n-2} h + \cdots + h^{n-1}) \)
\( = n x^{n-1} \cdots \) 答
\( (2) \)
\( f'(x) = \displaystyle\lim_{h \to 0} \dfrac{\sin (x+h) – \sin x}{h} \)
\( = \displaystyle\lim_{h \to 0} \dfrac{\sin x \cos h + \cos x \sin h – \sin x}{h} \)
\( = \displaystyle\lim_{h \to 0} \dfrac{\sin x (\cos h – 1) + \cos x \sin h}{h} \)
\( = \displaystyle\lim_{h \to 0} \left( \dfrac{(\cos h – 1)(\cos h + 1)}{\cos h + 1} \cdot \sin x + \dfrac{\sin h}{h} \cdot \cos x \right) \)
\( = \displaystyle\lim_{h \to 0} \left( \dfrac{- \sin^2 h}{\cos h + 1} \cdot \sin x + \dfrac{\sin h}{h} \cdot \cos x \right) \)
\( = \dfrac{0}{1+1} \cdot \sin x + 1 \cdot \cos x \)
\( = \cos x \cdots \) 答
\( (3) \)
\( f'(x) = \displaystyle\lim_{h \to 0} \dfrac{e^{x+h} – e^x}{h} \)
\( = \displaystyle\lim_{h \to 0} \dfrac{e^x (e^h – 1)}{h} \)
\( = e^x \cdot 1 \)
\( = e^x \cdots \) 答
\( (4) \)
\( f'(x) = \displaystyle\lim_{h \to 0} \dfrac{\log (x+h) – \log x}{h} \)
\( = \displaystyle\lim_{h \to 0} \dfrac{\log \left( 1 + \dfrac{h}{x} \right)}{h} \)
\( = \displaystyle\lim_{h \to 0} \dfrac{x}{h} \log \left( 1 + \dfrac{h}{x} \right) \cdot \dfrac{1}{x} \)
\( = \displaystyle\lim_{h \to 0} \log \left( 1 + \dfrac{h}{x} \right)^{\frac{x}{h}} \cdot \dfrac{1}{x} \)
\( = (\log e) \cdot \dfrac{1}{x} \)
\( = \dfrac{1}{x} \cdots \) 答
\( 1^\circ \)
\( (2) \) では
\( \displaystyle\lim_{h \to 0} \dfrac{\sin h}{h} = 1 \) ，
\( (3) \) では
\( \displaystyle\lim_{h \to 0} \dfrac{e^h – 1}{h} = 1 \) ，
\( (4) \) では
\( \displaystyle\lim_{t \to \infty} \left( 1 + \dfrac{1}{t} \right)^t = \displaystyle\lim_{h \to 0} (1+h)^{\frac{1}{h}} = e \)
となることをそれぞれ用いている．
