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### The Derivative of Arccosine

Cosine
##### Contents

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}}

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### March 16, 2018

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