# Derivative of Cosine Function/Proof 2

$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$
 $\ds \map {\frac \d {\d x} } {\cos x}$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\map \cos {x + h} - \cos x} h$ Definition of Derivative of Real Function at Point $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\cos x \cos h - \sin x \sin h - \cos x} h$ Cosine of Sum $\ds$ $=$ $\ds \lim_{h \mathop \to 0} \frac {\cos x \cos h - \cos x} h + \lim_{h \mathop \to 0} \frac {-\sin x \sin h} h$ Sum Rule for Limits of Real Functions $\ds$ $=$ $\ds \cos x \lim_{h \mathop \to 0} \frac {\cos h - 1} h - \sin x \lim_{h \mathop \to 0} \frac {\sin h} h$ Multiple Rule for Limits of Real Functions $\ds$ $=$ $\ds \cos x \times 0 - \sin x \times 1$ Limit of $\dfrac {\cos x - 1} x$ at Zero and Limit of $\dfrac {\sin x} x$ at Zero $\ds$ $=$ $\ds -\sin x$
$\blacksquare$