Derivative of Inverse Hyperbolic Tangent

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Theorem

Let $S$ denote the open real interval:

$S := \left({-1 \,.\,.\, 1}\right)$

Let $x \in S$.

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


Then:

$\dfrac \d {\d x} \left({\tanh^{-1} x}\right) = \dfrac 1 {1 - x^2}$


Proof

\(\displaystyle y\) \(=\) \(\displaystyle \tanh^{-1} x\)
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle \tanh y\) Definition of Inverse Hyperbolic Tangent
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d x} {\d y}\) \(=\) \(\displaystyle \operatorname{sech}^2 y\) Derivative of Hyperbolic Tangent Function
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d y} {\d x}\) \(=\) \(\displaystyle \frac 1 {\operatorname{sech}^2 y}\) Derivative of Inverse Function
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d y} {\d x}\) \(=\) \(\displaystyle \frac 1 {1 - \tanh^2 y}\) Sum of Squares of Hyperbolic Secant and Tangent
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d} {\d x} \left({\tanh^{-1} x}\right)\) \(=\) \(\displaystyle \frac 1 {1 - x^2}\) Definition of $x$ and $y$

$\blacksquare$


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