# Derivative of Inverse Hyperbolic Tangent

## 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

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

$\blacksquare$