Derivative of Inverse Hyperbolic Secant
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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