Derivative of Sine Function/Proof 2

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


\(\ds \map {\frac \d {\d x} } {\sin x}\) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\map \sin {x + h} - \sin x} h\) Definition of Derivative of Real Function at Point
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\sin x \cos h + \sin h \cos x - \sin x} h\) Sine of Sum
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\sin x \paren {\cos h - 1} + \sin h \cos x} h\) collecting terms containing $\map \sin x$ and factoring
\(\ds \) \(=\) \(\ds \lim_{h \mathop \to 0} \frac {\sin x \paren {\cos h - 1} } h + \lim_{h \mathop \to 0} \frac {\sin h \cos x} h\) Sum Rule for Limits of Real Functions
\(\ds \) \(=\) \(\ds \map \sin x \times 0 + 1 \times \cos x\) Limit of $\dfrac {\sin x} x$ at Zero and Limit of $\dfrac {\cos x - 1} x$ at Zero
\(\ds \) \(=\) \(\ds \cos x\)