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The Derivative of Arccosine


Cosine

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By a previous result

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

By convention,

{$$ \begin{gather} \cos(\arccos(\theta)) = \theta \cr 0 \le \theta \le \pi \end{gather} $$}


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

This demonstration proceeds by implicit differentiation.

{$$ \begin{align} \cos(\arccos(\theta)) &= \theta \tag{defintion of inverse} \cr \text{Let } y &= \arccos\theta \cr \cos(y) &= \theta \cr {d \over {d\theta}} \left( cos(y) \right) &= {d \over {d\theta}} \cr {d \over {d\theta}}\cos(y){d \over {d\theta}}y &= 1 \tag{Chain rule} \cr -\sin(y){d\over{d\theta}} &= 1 \tag{previous result} \cr {d\over{d\theta}} &= - {1 \over {\sin(y)}} \cr {d \over {d\theta}}y &= - {1 \over {\sqrt{1-\cos^2(y)}}} \tag{Pythagorean} \end{align} $$}

Since {$\sin(y)$} is non-negative in y's range, the positive root will do. To express the results in terms of θ:

{$$ \begin{align} {d \over {d\theta}}\arccos\theta &= - {1 \over {\sqrt(1+ cos^2(\arccos\theta))}} \cr \therefore \quad {d \over {d\theta}}\arccos\theta &= - {1 \over {\sqrt{1 - \theta^2}}}\end{align} $$}


Sources:

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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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August 05, 2017

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