定義に従って 1/x, √x を微分する（B）

(1) \[ \begin{aligned} f'(x)&=\lim_{h\to 0}\frac{1}{h}\left\{\frac{1}{(x+h)^2}-\frac{1}{x^2}\right\}=\lim_{h\to 0}\frac{x^2-(x+h)^2}{hx^2(x+h)^2}=\lim_{h\to 0}\frac{-h(2x+h)}{hx^2(x+h)^2}\\ &=\lim_{h\to 0}\frac{-2x-h}{x^2(x+h)^2}=\frac{-2x}{x^4}=-\frac{2}{x^3} \end{aligned} \] (2) $x>-\dfrac{1}{2}$ のとき \[ \begin{aligned} f'(x)&=\lim_{h\to 0}\frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}=\lim_{h\to 0}\frac{(2x+2h+1)-(2x+1)}{h(\sqrt{2x+2h+1}+\sqrt{2x+1})}\\ &=\lim_{h\to 0}\frac{2}{\sqrt{2x+2h+1}+\sqrt{2x+1}}=\frac{2}{2\sqrt{2x+1}}=\frac{1}{\sqrt{2x+1}} \end{aligned} \] (3) 加法定理より \[ \begin{aligned} f'(x)&=\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}=\lim_{h\to 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ &=\lim_{h\to 0}\left(\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}\right)=\sin x\cdot 0+\cos x\cdot 1=\cos x \end{aligned} \]