# Derivative of Inverse Hyperbolic Tangent

## 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 $y$

$\blacksquare$