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A previous result is: {$$ {d \over {d\theta}} \cot\theta = - \csc^2\theta $$}

Attention is called to the diagram which illustrates the Pythagorean identity:

{$$ csc^2\theta = 1 + cot^2\theta $$}

{$ y = \cot(x) $} and {$ y=\operatorname{arccot}(x) $}

The demonstration proceeds by implicit differentiation:

{$$ \begin{align} \cot(\operatorname{arccot}\theta) &= \theta \tag{definition of inverse} \cr \text{Let } y &= \operatorname{arccot}\theta \cr \cot(y) &= \theta \cr {d \over {d\theta}} \left( \cot(y) \right) &= {d \over {d\theta}} \theta \cr {d \over {d\theta}}cot(y){d \over {d\theta}}y &= 1 \tag{Chain rule} \cr -\csc^2{d \over {d\theta}}y &= 1 \tag{previous result} \cr {d \over {d\theta}}y &= - {1 \over {csc^2(y)}} \cr {d \over {d\theta}}y & = - {1 \over {1 + cot^2(y)}} \tag{Pythagorean identity} \cr {d \over {d\theta}}\operatorname{arccot}\theta &= - {1 \over {1 + cot^2(\operatorname{arccot}\theta)}} \tag{value y} \cr \therefore \quad {d \over {d\theta}}\operatorname{arccot}\theta &= - {1 \over {1 + \theta^2}}\end{align} $$}

{$ y = \cot(x) $}, {$ y=\operatorname{arccot}(x)$}, {$ y = \cot(x) $} and {$ y=\operatorname{arccot}(x) $}.

*Sources:*

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**Category:** Math Calculus Trigonometry

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