• Algebra
• Trigonometry
• Geometry
• Calculus

Inverse Hyperbolic Tangent function

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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