与えられた3つの関数について、導関数の定義に従って微分を求めます。 (1) $f(x) = \frac{1}{x-2}$ (2) $f(x) = \frac{1}{x^2}$ (3) $f(x) = \sqrt{2x}$

1. 問題の内容

与えられた3つの関数について、導関数の定義に従って微分を求めます。

(1) $f(x) = \frac{1}{x-2}$

(2) $f(x) = \frac{1}{x^2}$

(3) $f(x) = \sqrt{2x}$

2. 解き方の手順

導関数の定義は、

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

です。

(1) $f(x) = \frac{1}{x-2}$ の場合

$f(x+h) = \frac{1}{x+h-2}$

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h-2} - \frac{1}{x-2}}{h}$

$f'(x) = \lim_{h \to 0} \frac{(x-2) - (x+h-2)}{h(x+h-2)(x-2)}$

$f'(x) = \lim_{h \to 0} \frac{-h}{h(x+h-2)(x-2)}$

$f'(x) = \lim_{h \to 0} \frac{-1}{(x+h-2)(x-2)}$

$f'(x) = \frac{-1}{(x-2)^2}$

(2) $f(x) = \frac{1}{x^2}$ の場合

$f(x+h) = \frac{1}{(x+h)^2}$

$f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$

$f'(x) = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{h(x+h)^2 x^2}$

$f'(x) = \lim_{h \to 0} \frac{x^2 - (x^2 + 2xh + h^2)}{h(x+h)^2 x^2}$

$f'(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h(x+h)^2 x^2}$

$f'(x) = \lim_{h \to 0} \frac{-2x - h}{(x+h)^2 x^2}$

$f'(x) = \frac{-2x}{x^4} = \frac{-2}{x^3}$

(3) $f(x) = \sqrt{2x}$ の場合

$f(x+h) = \sqrt{2(x+h)} = \sqrt{2x+2h}$

$f'(x) = \lim_{h \to 0} \frac{\sqrt{2x+2h} - \sqrt{2x}}{h}$

$f'(x) = \lim_{h \to 0} \frac{(\sqrt{2x+2h} - \sqrt{2x})(\sqrt{2x+2h} + \sqrt{2x})}{h(\sqrt{2x+2h} + \sqrt{2x})}$

$f'(x) = \lim_{h \to 0} \frac{2x+2h - 2x}{h(\sqrt{2x+2h} + \sqrt{2x})}$

$f'(x) = \lim_{h \to 0} \frac{2h}{h(\sqrt{2x+2h} + \sqrt{2x})}$

$f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2x+2h} + \sqrt{2x}}$

$f'(x) = \frac{2}{\sqrt{2x} + \sqrt{2x}} = \frac{2}{2\sqrt{2x}} = \frac{1}{\sqrt{2x}}$

3. 最終的な答え

(1) $f'(x) = \frac{-1}{(x-2)^2}$

(2) $f'(x) = \frac{-2}{x^3}$

(3) $f'(x) = \frac{1}{\sqrt{2x}}$