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The thought that since cotangent is the reciprocal of tangent, its derivative might be the reciprocal of the derivative of tangent is not quite right.

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

{$$ \begin{align} {d \over {d\theta}} \cot\theta &= {d \over {d\theta}} \left( {{\cos\theta} \over {\sin\theta}} \right) \cr &= {\sin\theta{d \over {d\theta}}\cos\theta - \cos\theta{d \over{d\theta}}\sin\theta \over {\sin^2\theta}} \tag{Quotient rule} \cr &= {{-\sin^2\theta - \cos2\theta} \over {\sin^2\theta}} \cr &= - \left( {{\sin^2\theta} \over {\sin^2\theta}} + {{\cos2\theta} \over {\sin^2\theta}} \right) \cr &= - \left( 1 + \cot^2\theta \right) \cr \therefore \quad {d \over {d\theta}} \cot\theta &= - \csc^2\theta \tag{Pythagorean identity} \end{align} $$}

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**Category:** Math Calculus Trigonometry

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