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The Derivative of Cosine

Cosine

{$ y = \cos(x) $} and {$ y=\arccos(x) $}

{$$ \begin{gather} {d \over {dx}} \cos x = \lim_{h \rightarrow 0} {{ \cos(x+h) - \cos x} \over h} \cr = \lim_{h \rightarrow 0} {{ \cos(x) \cos(h) - \sin (x) \sin (h) - \cos(x)} \over h} \cr = \lim_{h \rightarrow 0} {{ \cos(x)( \cos(h) - 1) - \sin (x) \sin (h)} \over h} \cr = \lim_{h \rightarrow 0} \left[ \cos(x) \left( {{\cos(h) - 1} \over h} \right) - \sin (x) \left( {{\sin (h)} \over h} \right) \right] \end{gather} $$}

The limit of {$(\cos(h)-1)/h$} is known to be 0 from a previous result, and likewise {$\sin(x)/h$} is known to be 1. Therefore,

{$$ {d \over {dx}} \cos x = - \sin x $$}

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Category: Math Calculus

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December 23, 2018