# 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 $\sech^{-1} x$ denote the inverse hyperbolic secant of $x$.

Then:

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

## Proof

$\sech^{-1} x$ is defined only on the half-open real interval $\hointl 0 1$.

Thus on $\hointl 0 1$:

\(\displaystyle y\) | \(=\) | \(\displaystyle \sech^{-1} x\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle \sech y\) | where $y \in \R_{>0}$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d x} {\d y}\) | \(=\) | \(\displaystyle -\sech y \, \tanh y\) | Derivative of Hyperbolic Secant Function | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d y} {\d x}\) | \(=\) | \(\displaystyle \dfrac {-1} {\sech y \ \tanh y}\) | Derivative of Inverse Function | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d y} {\d x}\) | \(=\) | \(\displaystyle \frac {-1} {\sech y \, \sqrt {1 - \sech^2 y} }\) | Sum of Squares of Hyperbolic Secant and Tangent | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {\frac \d {\d x} } {\sech^{-1} x}\) | \(=\) | \(\displaystyle \frac {-1} {x \sqrt {1 - x^2} }\) | 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 $\sech^{-1} x$ can be defined only on $\openint 0 1$.

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.41$