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Derivative of Arctan Demonstration


Arctan

A previous result is

{$$ {d \over {d\theta}} \tan\theta = \sec^2\theta $$}

By convention {$\arctan(y)$} is defined for:

{$$ -{\pi \over 2} \lt y \lt {\pi \over 2} $$}

or in other words, in the first and second quadrant.


{$ \color{forestgreen}{y = \tan(x)} $}, {$ \color{crimson}{y=\arctan(x)} $}, and {$ \color{blue}{y= d/dx \arctan(x)} $}

The demonstration proceeds by implicit differentiation.

{$$ \large \begin{align} \tan(\arctan\theta) &= \theta \tag{definition of inverse} \cr \text{Let } y &= \arctan\theta \cr \tan(y) &= \theta \cr {d \over {d\theta}}\left( \tan(y) \right) &= {d \over {d\theta}}\theta \cr \sec^2(\theta){d \over {d\theta}}y &= 1 \tag{Chain rule}\cr {d \over {d\theta}} &= {1 \over {sec^2(\theta)}} \cr {d \over {d\theta}} = {1 \over {1 + tan^2(\theta)}} \tag{Pythagorean identity} \cr {d \over {d\theta}}\arctan\theta &= {1 \over {1+ tan^2(\arctan\theta)}} \end{align} $$} {$$ \Large \therefore \quad {d \over {d\theta}}arctan\theta = {1 \over {1 + \theta^2}} $$}


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