Derivative of Inverse Hyperbolic Tangent

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Theorem

Let $S$ denote the open real interval:

$S := \openint {-1} 1$

Let $x \in S$.

Let $\tanh^{-1} x$ be the inverse hyperbolic tangent of $x$.


Then:

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


Proof

\(\ds y\) \(=\) \(\ds \tanh^{-1} x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \tanh y\) Definition of Real Inverse Hyperbolic Tangent
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d y}\) \(=\) \(\ds \sech^2 y\) Derivative of Hyperbolic Tangent
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac 1 {\sech^2 y}\) Derivative of Inverse Function
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac 1 {1 - \tanh^2 y}\) Sum of Squares of Hyperbolic Secant and Tangent
\(\ds \leadsto \ \ \) \(\ds \map {\frac \d {\d x} } {\tanh^{-1} x}\) \(=\) \(\ds \frac 1 {1 - x^2}\) Definition of $x$ and $y$

$\blacksquare$


Sources