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### The Derivative of Sine

##### Contents

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

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,

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

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Category: Math Calculus

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

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### October 21, 2018

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