# Derivative of Sine Function/Proof 4

## Theorem

$\map {\dfrac \d {\d x} } {\sin x} = \cos x$

## Proof

 $\ds \map {\frac \d {\d x} } {\sin x}$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h$ Definition of Derivative of Real Function at Point $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map \sin {\paren {x + \frac h 2} + \frac h 2} - \map \sin {\paren {x + \frac h 2} - \frac h 2} } h$ $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {2 \map \cos {x + \frac h 2} \map \sin {\frac h 2} } h$ Simpson's Formula for Cosine by Sine $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \map \cos {x + \frac h 2} \lim_{h \mathop \to 0} \frac {\map \sin {\frac h 2} } {\frac h 2}$ Multiple Rule for Limits of Real Functions and Product Rule for Limits of Real Functions $\ds$ $=$ $\ds \cos x \times 1$ Cosine Function is Continuous and Limit of $\dfrac {\sin x} x$ at Zero $\ds$ $=$ $\ds \cos x$

$\blacksquare$