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Establish the formula for the derivative of arccsc x

by RoRi

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

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

For x > 1 let

    \[ y = \operatorname{arccsc} x \quad \implies \quad x = \csc y. \]

Then we know

    \[ \frac{dx}{dy} = -\csc y \cot y. \]

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

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

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

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

2 comments

    • Anonymous says:

      We can take the positive or negative square root. However, the restricted csc function, for which arccsc is the inverse, is decreasing everywhere it is defined. The inverse of a decreasing function is decreasing. So arccsc is decreasing everywhere, and its derivative must be negative everywhere.

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