Derivative of Sine Function/Proof 4

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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
\(\ds \) \(=\) \(\ds \cos x\)

$\blacksquare$