sec 미분 증명 ¶
$\displaystyle \frac{d}{dx}(\sec x)$ 은 미분의 정의에 따라
$\displaystyle =\lim_{h\to0}\frac{\sec(x+h)-\sec x}{h}$
sec(x)의 역수는 cos(x)이므로,$\displaystyle =\lim_{h\to0}\frac{\frac1{\cos(x+h)}-\frac1{\cos x}}{h}$
$\displaystyle =\lim_{h\to0}\frac{\frac{\cos x}{\cos x\cdot\cos(x+h)}-\frac{\cos(x+h)}{\cos x\cdot\cos(x+h)}}{h}$
$\displaystyle =\lim_{h\to0}\frac{\frac{\cos x-\cos(x+h)}{\cos x\cdot\cos(x+h)}}{\frac{h}{1}}$
$\displaystyle =\lim_{h\to0}\frac{\cos x-\cos(x+h)}{h\cdot \cos x\cdot\cos(x+h)}$
코사인의 차 공식에 의해 $\displaystyle \cos(x+h)=\cos x\cos h-\sin x\sin h$ 이고 이것을 분자에만 적용하면,$\displaystyle =\lim_{h\to0}\frac{\frac{\cos x}{\cos x\cdot\cos(x+h)}-\frac{\cos(x+h)}{\cos x\cdot\cos(x+h)}}{h}$
$\displaystyle =\lim_{h\to0}\frac{\frac{\cos x-\cos(x+h)}{\cos x\cdot\cos(x+h)}}{\frac{h}{1}}$
$\displaystyle =\lim_{h\to0}\frac{\cos x-\cos(x+h)}{h\cdot \cos x\cdot\cos(x+h)}$
$\displaystyle =\lim_{h\to0}\frac{\cos x-\cos x\cos h+\sin x\sin h}{h\cdot\cos x\cdot\cos(x+h)}$
$\displaystyle =\lim_{h\to0}\left[\frac{\cos x(1-\cos h)+\sin x\cdot\sin h}{h\cdot\cos x\cdot\cos(x+h)}\right]$
$\displaystyle =\lim_{h\to0}\left[\frac{1-\cos h}{h\cdot\cos(x+h)}+\frac{\sin x\cdot\sin h}{h\cdot\cos x\cdot\cos(x+h)}\right]$
$\displaystyle =\lim_{h\to0}\left[\frac{1-\cos h}{h}\cdot\frac{1}{\cos(x+h)}+\frac{\sin h}{h}\cdot\frac{\sin x}{\cos x\cdot\cos(x+h)}\right]$
$\displaystyle =\lim_{h\to0}\left[\frac{1-\cos h}{h}\cdot\frac{1}{\cos(x+h)}+\frac{\sin h}{h}\cdot\frac{1}{\cos x}\cdot\frac{\sin x}{\cos(x+h)}\right]$
$\displaystyle =0\cdot\sec x+1\cdot\sec x\cdot\tan x$
$\displaystyle =\sec x\tan x$
$\displaystyle =\lim_{h\to0}\left[\frac{\cos x(1-\cos h)+\sin x\cdot\sin h}{h\cdot\cos x\cdot\cos(x+h)}\right]$
$\displaystyle =\lim_{h\to0}\left[\frac{1-\cos h}{h\cdot\cos(x+h)}+\frac{\sin x\cdot\sin h}{h\cdot\cos x\cdot\cos(x+h)}\right]$
$\displaystyle =\lim_{h\to0}\left[\frac{1-\cos h}{h}\cdot\frac{1}{\cos(x+h)}+\frac{\sin h}{h}\cdot\frac{\sin x}{\cos x\cdot\cos(x+h)}\right]$
$\displaystyle =\lim_{h\to0}\left[\frac{1-\cos h}{h}\cdot\frac{1}{\cos(x+h)}+\frac{\sin h}{h}\cdot\frac{1}{\cos x}\cdot\frac{\sin x}{\cos(x+h)}\right]$
$\displaystyle =0\cdot\sec x+1\cdot\sec x\cdot\tan x$
$\displaystyle =\sec x\tan x$
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