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Establish the formula for the derivative of arccot x

by RoRi

Establish the following formula for the derivative of \arctan x is correct,

    \[ D(\operatorname{arccot} x) = \frac{-1}{1+x^2}, \qquad \text{for all } x \in \mathbb{R}. \]

For x \in \mathbb{R} let

    \[ y = \operatorname{arccot} x \quad \implies \quad x = \cot y. \]

Then we know

    \[ \frac{dx}{dy} = -\csc^2 y. \]

Therefore, by Theorem 6.7 (p. 252 of Apostol) we have,

    \begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && -\csc^2 y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{-1}{\csc^2 y} \\ && \implies && D(\operatorname{arccot} x) &= \frac{-1}{\csc^2 y}. \end{align*}

Using a trig identity for tangent and cosecant we have

    \[ \csc^2 y = 1 + \cot^2 y = 1 + x^2 \]

since x = \cot y. Therefore we conclude,

    \[ D(\operatorname{arccot} x) = \frac{-1}{1+x^2}.\]

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