Derivative of Inverse Hyperbolic Secant

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S$ denote the open real interval:

$S := \openint 0 1$

Let $x \in S$.

Arsech.png

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 $\openint 0 1$.




$\blacksquare$


Sources