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