Derivative of Sine

Let $y=f(x) = \sin x $, then

$  {{dy} \over {dx}} = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$

$  = \mathop {\lim }\limits_{h \to 0} {{\sin (x + h) - \sin x} \over h}$

$  = \mathop {\lim }\limits_{h \to 0} {{\sin x\cos h  + \cos x\sin h  - \sin x} \over h}$

$  = \mathop {\lim }\limits_{h \to 0} {{\sin x(\cos h  - 1) + \cos x\sin h } \over h}$

$  = \mathop {\lim }\limits_{h \to 0} {{\sin x(\cos h  - 1)} \over h} + \mathop {\lim }\limits_{h \to 0} {{\cos x\sin h } \over h}$

$  = (\sin x) \mathop {\lim }\limits_{h \to 0} {{\cos h  - 1} \over h} + (\cos x) \mathop {\lim }\limits_{h \to 0} {{\sin h } \over h}$

$  = (\sin x)(0) + (\cos x)(1)$

$  = \cos x$

In the above, we have used the fact that $  \mathop {\lim }\limits_{h \to 0} {{\cos h  - 1} \over h} = 0$ and $  \mathop {\lim }\limits_{h \to 0} {{\sin h } \over h} = 1$