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Establish the formula for the derivative of arcsec x

by RoRi

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

    \[ D(\operatorname{arcsec} x) = \frac{1}{|x|\sqrt{x^2-1}}, \qquad \text{for all } |x| >1. \]

For x > 1 let

    \[ y = \operatorname{arcsec} x \quad \implies \quad x = \sec y. \]

Then we know

    \[ \frac{dx}{dy} = \sec y \tan y. \]

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

    \begin{align*} \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} && \implies && \sec y \tan y &= \frac{1}{\frac{dy}{dx}} \\ && \implies && \frac{dy}{dx} &= \frac{1}{\sec y \tan y} \\ && \implies && D(\operatorname{arcsec} x) &= \frac{1}{\sec y \tan y} \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \left( \frac{1}{\sec y \tan y} \right)^2 \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \frac{1}{\sec^2 y \tan^2 y} \\ && \implies && (D(\operatorname{arcsec} x))^2 &= \frac{1}{\sec^2 y (\sec^2 y - 1)}. \end{align*}

Where we use the trig identity \tan^2 y = \sec^2 y - 1 in the final line. Then, since x = \sec y we have,

    \begin{align*} &&(D(\operatorname{arcsec} x))^2 &= \frac{1}{x^2 (x^2 - 1)} \\ \implies && D (\operatorname{arcsec} x) &= \frac{1}{|x|\sqrt{x^2-1}}. \qquad \blacksquare \end{align*}

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