# Inverse Hyperbolic Tangent function

The inverse form of the hyperbolic tangent function is called the inverse hyperbolic tangent function.

## Formula

$\large \tanh^{-1}{x} \,=\, \dfrac{1}{2} \, \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}$

The hyperbolic tangent function is defined in mathematics as the ratio of subtraction to summation of negative and positive natural exponential functions. The inverse form of the hyperbolic tangent function is in logarithmic function form and it can be derived from the hyperbolic tangent function in mathematics.

### Proof

$x$ and $y$ are two literals. The value of $x$ is equal to the hyperbolic tangent of $y$.

$x = \tanh{y}$

Therefore, the value of $y$ is equal to the inverse hyperbolic tangent of $x$.

$y = \tanh^{-1}{x}$

So, the relation between them can be written in mathematics as follows.

$x = \tanh{y} \,\,\Leftrightarrow \,\, y = \tanh^{-1}{x}$

01.

#### Transforming Hyperbolic function in exponential form

Express hyperbolic tangent of $y$ in terms of natural exponential functions.

$x = \dfrac{e^y-e^{-y}}{e^y+e^{-y}}$

02.

#### Simplification of the equation

Simplify the exponential algebraic equation to express the equation in terms of $x$ and also to eliminate the $y$ from the equation.

$\implies$ $x(e^y+e^{-y}) = e^y-e^{-y}$

$\implies$ $xe^y+xe^{-y} = e^y-e^{-y}$

$\implies$ $0 = e^y-e^{-y}-xe^y-xe^{-y}$

$\implies$ $e^y-e^{-y}-xe^y-xe^{-y} = 0$

$\implies$ $e^y-xe^y-e^{-y}-xe^{-y} = 0$

$\implies$ $(1-x)e^y-(1+x)e^{-y} = 0$

$\implies$ $(1-x)e^y = (1+x)e^{-y}$

$\implies$ $(1-x)e^y = \dfrac{1+x}{e^y}$

$\implies$ $(1-x)e^y \times e^y = 1+x$

$\implies$ $(1-x)e^{2y} = 1+x$

$\implies$ $e^{2y} = \dfrac{1+x}{1-x}$

03.

#### Eliminating the y from equation

Eliminate $y$ from this algebraic exponential equation. It is possible by applying natural logarithm both sides of the equation.

$\implies$ $\log_{e}{e^{2y}} = \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}$

$\implies$ $2y \times \log_{e}{e} = \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}$

$\implies$ $2y \times 1 = \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}$

$\implies$ $2y = \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}$

$\implies$ $y = \dfrac{1}{2} \, \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\tanh^{-1}{x} = \dfrac{1}{2} \, \log_{e}{\Bigg(\dfrac{1+x}{1-x}\Bigg)}$

Latest Math Topics
Jun 26, 2023
Jun 23, 2023

Latest Math Problems
Jul 01, 2023
Jun 25, 2023
###### Math Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Practice now

###### Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

###### Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.