= Theorem =
 $\frac{d}{dx}\cos x=-\sin x$

= Proof 1 =
식
 $\frac{\cos(x+h)-\cos x}{h}$
 $=\frac{\cos x\cos h-\sin x\sin h-cos x}{h}$
 $=\cos x\left(\frac{\cos h-1}{h}\right)-\sin x\left(\frac{\sin h}{h}\right)$
$h\to 0$ 일 때의 극한을 구하면
 $=\cos x\cdot 0 -\sin x \cdot 1$
 $=-\sin x$

i.e.

Let $f(x)=\cos x$
$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
 $=\lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h}$
 $=\lim_{h\to 0}\frac{\cos x\cdot\cos h-\sin x\cdot\sin h-\cos x}{h}$
 $=\lim_{h\to 0}\frac{\cos x(\cos h-1)-\sin x\cdot\sin h}{h}$
 $=\lim_{h\to 0}\left(\cos x\frac{\cos h-1}{h}-\sin x\cdot\frac{\sin h}{h}\right)$
좌측에 [[VG:로피탈_정리,L_Hopital_s_rule]]를 쓰면
 $=\cos x\cdot 0 - \sin x \cdot 1$
 $=-\sin x$
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$\frac{d}{dx}(\cos x)$ 는,
 $=\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}$
 $=\lim_{h\to0}\frac{\cos x\cdot\cos h-\sin x\cdot\sin h-\cos x}{h}$
 $=\lim_{h\to0}\frac{\cos x\cdot\cos h-\cos x-\sin x\cdot\sin h}{h}$
 $=\lim_{h\to0}\frac{\cos x\cdot(\cos h-1)}{h}-\lim_{h\to0}\frac{\sin x\cdot\sin h}{h}$
 $=\cos x\cdot\lim_{h\to0}\frac{\cos h-1}{h}-\sin x\cdot\lim_{h\to0}\frac{\sin h}{h}$
 $=\cos x\cdot 0-\sin x\cdot 1$
 $=-\sin x$

= Proof 2 =
Let $f(x)=\cos x$
$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
 $=\lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h}$
[[VG:삼각함수,trigonometric_function#s-8.2]] 합·차 → 곱 공식을 쓰면,
 $=\lim_{h\to 0}\left(-2\sin\frac{x+h+x}{2}\cdot\sin\frac{x+h-x}{2}\right)/h$
 $=\lim_{h\to 0}\left(-2\sin\frac{2x+h}{2}\cdot\sin\frac{h}{2}\right)/h$
 $=\lim_{h\to 0}\frac{-\sin\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{\frac{h}{2}}$
 $=-\sin(x+0)=-\sin x=f'(x)$

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코사인 미분 증명
from [[VG:삼각함수_미분표]]

Up: [[여러가지증명]]