Derivative of Inverse Hyperbolic Cotangent

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Theorem

Let $S$ denote the union of the unbounded open real intervals:

$S := \openint \gets {-1} \cup \openint 1 \to$

Let $x \in S$.

Let $\coth^{-1} x$ be the inverse hyperbolic cotangent of $x$.


Then:

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


Proof

\(\ds y\) \(=\) \(\ds \coth^{-1} x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \coth y\) Definition of Real Inverse Hyperbolic Cotangent
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d y}\) \(=\) \(\ds -\csch^2 y\) Derivative of Hyperbolic Cotangent
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac {-1} {\csch^2 y}\) Derivative of Inverse Function
\(\ds \) \(=\) \(\ds \frac {-1} {\coth^2 y - 1}\) Difference of Squares of Hyperbolic Cotangent and Cosecant
\(\ds \leadsto \ \ \) \(\ds \map {\frac \d {\d x} } {\coth^{-1} x}\) \(=\) \(\ds \frac {-1} {x^2 - 1}\) Definition of $x$ and $y$

$\blacksquare$


Also presented as

This result can also be (and usually is) reported as:

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

but this obscures the fact that $x^2 > 1$ in order for it to be defined.


Sources

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