Derivative of Inverse Hyperbolic Secant

Theorem

Let $S$ denote the open real interval:

$S := \openint 0 1$

Let $x \in S$.

Let $\arsech x$ denote the real inverse hyperbolic secant of $x$.

Then:

$\map {\dfrac \d {\d x} } {\arsech x} = \dfrac {-1} {x \sqrt{1 - x^2} }$

Proof

$\arsech x$ is defined only on the half-open real interval $\hointl 0 1$.

Thus on $\hointl 0 1$:

$\ds y$

$=$

$\ds \arsech x$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds \sech y$

where $y \in \R_{>0}$

$\ds \leadsto \ \ $

$\ds \frac {\d x} {\d y}$

$=$

$\ds -\sech y \tanh y$

Derivative of Hyperbolic Secant

$\ds \leadsto \ \ $

$\ds \frac {\d y} {\d x}$

$=$

$\ds \dfrac {-1} {\sech y \ \tanh y}$

Derivative of Inverse Function

$\ds \leadsto \ \ $

$\ds \frac {\d y} {\d x}$

$=$

$\ds \frac {-1} {\sech y \sqrt {1 - \sech^2 y} }$

Sum of Squares of Hyperbolic Secant and Tangent

$\ds \leadsto \ \ $

$\ds \map {\frac \d {\d x} } {\arsech x}$

$=$

$\ds \frac {-1} {x \sqrt {1 - x^2} }$

Definition of $x$ and $y$

When $x = 1$, however, $\sqrt{1 - x^2} = 0$ and so $\dfrac {-1} {x \sqrt {1 - x^2} }$ is undefined.

Hence $\arsech x$ can be defined only on $\openint 0 1$.

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$\blacksquare$

Sources

1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Appendix $2$: Table of derivatives and integrals of common functions: Inverse hyperbolic functions
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Appendix: Table $1$: Derivatives
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Appendix: Table $1$: Derivatives