사인 함수 미분 증명
Proof 1 ¶
Let $\displaystyle f(x)=\sin x$
$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
로피탈_정리,L_Hopital_s_rule를 쓰면
$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\sin x\cdot\cos h+\cos x\cdot\sin h-\sin x}{h}$
$\displaystyle =\lim_{h\to 0}\left(\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}\right)$
$\displaystyle =\sin x\cdot\lim_{h\to 0}\frac{\cos h-1}{h}+\cos x\cdot\lim_{h\to 0}\frac{\sin h}{h}$
좌측에 $\displaystyle =\lim_{h\to 0}\frac{\sin x\cdot\cos h+\cos x\cdot\sin h-\sin x}{h}$
$\displaystyle =\lim_{h\to 0}\left(\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}\right)$
$\displaystyle =\sin x\cdot\lim_{h\to 0}\frac{\cos h-1}{h}+\cos x\cdot\lim_{h\to 0}\frac{\sin h}{h}$
로피탈_정리,L_Hopital_s_rule를 쓰면$\displaystyle =\sin x\cdot 0+\cos x\cdot 1$
$\displaystyle =\cos x$
i.e.$\displaystyle =\cos x$
$\displaystyle \frac{\sin(x+h)-\sin x}{h}$
$\displaystyle =\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$
$\displaystyle =\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right)$
$\displaystyle h\to 0$ 일 때의 극한을 구하면$\displaystyle =\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$
$\displaystyle =\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right)$
$\displaystyle =\sin x \cdot 0 + \cos x \cdot 1$
$\displaystyle =\cos x$
$\displaystyle =\cos x$
Proof 2 ¶
Let $\displaystyle f(x)=\sin x$
삼각함수,trigonometric_function#s-8.2 합·차 → 곱 공식을 쓰면,
$\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$
삼각함수,trigonometric_function#s-8.2 합·차 → 곱 공식을 쓰면,$\displaystyle =\lim_{h\to 0}\left(2\cos\frac{x+h+x}{2}\cdot\sin\frac{x+h-x}{2}\right)/h$
$\displaystyle =\lim_{h\to 0}\left(2\cos\frac{2x+h}{2}\cdot\sin\frac{h}{2}\right)/h$
$\displaystyle =\lim_{h\to 0}\frac{2\cos\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\cos\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{\frac{h}{2}}$
$\displaystyle =\cos(x+0)\cdot 1$
$\displaystyle =\cos x$
$\displaystyle =\lim_{h\to 0}\left(2\cos\frac{2x+h}{2}\cdot\sin\frac{h}{2}\right)/h$
$\displaystyle =\lim_{h\to 0}\frac{2\cos\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\cos\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{\frac{h}{2}}$
$\displaystyle =\cos(x+0)\cdot 1$
$\displaystyle =\cos x$
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