# Find the derivative of $\cos\: x$ from first principle.

Let $f(x) = \cos\: x$ from first principle.
$f'(x) = \lim\limits_{h \to 0} \large\frac{f(x+h)-f(x)}{h}$
$= \lim\limits_{h \to 0} \large\frac{\cos (x+h)-\cos (x)}{h}$
$= \lim\limits_{h \to 0} \bigg[ \large\frac{\cos x \: \cos h-\sin x \: \sin h - \cos x }{h}\bigg]$
$= \lim\limits_{h \to 0} \bigg[ \large\frac{-\cos x (1- \cos h)-\sin x \: \sin h }{h}\bigg]$
$= \lim\limits_{h \to 0} \bigg[ \large\frac{-\cos x (1- \cos h) }{h}$$-\Large\frac{-\sin x \: \sin h}{h}\bigg] = - \cos x \bigg( \lim\limits_{h \to 0} \large\frac{1- \cos h}{h} \bigg)$$-\sin x \lim\limits_{h \to 0} \bigg( \large\frac{\sin h}{h} \bigg)$
$= -\cos x(0)- \sin x (1) \qquad \bigg[ \lim\limits_{h \to 0} \large\frac{1-\cos h}{h}$$=0 \: and \: \lim\limits_{h \to 0} \large\frac{\sin h}{h}=1 \bigg]$
$= - \sin x$
$\therefore f'(x) = - \sin x$