Let $y=f(x) = \cos x $, then
${{dy} \over {dx}} = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$
$ = \mathop {\lim }\limits_{h \to 0} {{\cos (x + h) - \cos x} \over h}$
$ = \mathop {\lim }\limits_{h \to 0} {{\cos x\cos h - \sin x\sin h - \cos x} \over h}$
$ = \mathop {\lim }\limits_{h \to 0} {{\cos x(\cos h - 1) - \sin x\sin h } \over h}$
$ = \mathop {\lim }\limits_{h \to 0} {{\cos x(\cos h - 1)} \over h} - \mathop {\lim }\limits_{h \to 0} {{\sin x\sin h } \over h}$
$ = \cos x\mathop {\lim }\limits_{h \to 0} {{\cosh - 1} \over h} - \sin x\mathop {\lim }\limits_{h \to 0} {{\sin h } \over h}$
$ = (\cos )(0) - (\sin x)(1)$
$ = - \sin x$
In the above, we have used the fact that $ \displaystyle \mathop {\lim }\limits_{h \to 0} {{\cos h - 1} \over h} = 0$ and $ \displaystyle \mathop {\lim }\limits_{h \to 0} {{\sin h } \over h} = 1$
