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Establish the formula for the derivative of arccos x

by RoRi

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

    \[ D(\arccos x) = \frac{-1}{\sqrt{1-x^2}}, \qquad \text{if } -1<x<1. \]

For -1<x<1 let

    \[ y = \arccos x \quad \implies \quad x = \cos y. \]

Then we know

    \[ \frac{dx}{dy} = -\sin y. \]

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

    \begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && -\sin y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{-1}{\sin y} \\ && \implies && D(\arccos x) &= \frac{-1}{\sin y}. \end{align*}

Using the pythagorean identity for sine and cosine we have

    \begin{align*} \sin^2 y + \cos^2 y =1 && \implies && \sin y &= \pm \sqrt{1-\cos^2 y} \\ && \implies && \sin y &= \pm \sqrt{1-x^2} \end{align*}

since x = \cos y. Furthermore, since \sin y \geq 0 on the range of \arccos x (i.e., \sin y \ge 0 for y \in [0, \pi]) we must take the positive square root. Therefore we conclude,

    \[ D(\arccos x) = \frac{-1}{\sqrt{1-x^2}}.\]

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