Find the derivative of cos x from first principle.

Let f (x) = cos x. Accordingly, from the first principle,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h}$

$=\lim _{h \rightarrow 0}\left[\frac{\cos x \cos h-\sin x \sin h-\cos x}{h}\right]$

$=\lim _{h \rightarrow 0}\left[\frac{-\cos x(1-\cos h)-\sin x \sin h}{h}\right]$

$=\lim _{h \rightarrow 0}\left[\frac{-\cos x(1-\cos h)}{h}-\frac{\sin x \sin h}{h}\right]$

$=-\cos x\left(\lim _{h \rightarrow 0} \frac{1-\cos h}{h}\right)-\sin x \lim _{h \rightarrow 0}\left(\frac{\sin h}{h}\right)$

$=-\cos x(0)-\sin x(1)$ $\left[\lim _{h \rightarrow 0} \frac{1-\cos h}{h}=0\right.$ and $\left.\lim _{h \rightarrow 0} \frac{\sin h}{h}=1\right]$

$=-\sin x$

$\therefore f^{\prime}(x)=-\sin x$