Question

How do you show that ${{\tanh }^{-1}}\left( x \right)=\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x} \right)$?

Answer 1

We first describe the concept of hyperbolic functions. We express the basic formula of $\tanh y=\dfrac{{{e}^{2y}}-1}{{{e}^{2y}}+1}$. We used the inverse formula and then tried to find the value of $y$ with respect to $x$. Then using the logarithm formula we prove the given expression.

Complete step by step solution:
We use the concept of hyperbolic functions and formulas of logarithms.
We have been given the expression of ${{\tanh }^{-1}}\left( x \right)$. We assume ${{\tanh }^{-1}}\left( x \right)=y$.
Using the concept of inverse law, we get $x=\tanh y$.
We have the hyperbolic equation formula where $\tanh y=\dfrac{{{e}^{2y}}-1}{{{e}^{2y}}+1}$.
Therefore, the equation becomes $x=\tanh y=\dfrac{{{e}^{2y}}-1}{{{e}^{2y}}+1}$. We have $x=\dfrac{{{e}^{2y}}-1}{{{e}^{2y}}+1}$.
We try to find the value of $y$ with respect to $x$.
We apply the componendo-dividendo formula on the equation $\dfrac{x}{1}=\dfrac{{{e}^{2y}}-1}{{{e}^{2y}}+1}$ to find
$\begin{aligned} & \dfrac{x}{1}=\dfrac{{{e}^{2y}}-1}{{{e}^{2y}}+1} \\\ & \Rightarrow \dfrac{1}{x}=\dfrac{{{e}^{2y}}+1}{{{e}^{2y}}-1} \\\ & \Rightarrow \dfrac{1+x}{1-x}=\dfrac{{{e}^{2y}}+1+{{e}^{2y}}-1}{{{e}^{2y}}+1-{{e}^{2y}}+1}=\dfrac{2{{e}^{2y}}}{2}={{e}^{2y}} \\\ & \Rightarrow {{e}^{2y}}=\dfrac{1+x}{1-x} \\\ \end{aligned}$
We now apply the logarithm formula to find the value of $y$.
We have $\log {{x}^{a}}=a\log x$. The power value of $a$ goes as a multiplication with $\log x$.
In case of logarithmic numbers having powers, we have to multiply the power in front of the logarithm to get the single logarithmic function. We have $\ln a={{\log }_{e}}a$.
Therefore, ${{\log }_{e}}\left( {{e}^{2y}} \right)={{\log }_{e}}\left( \dfrac{1+x}{1-x} \right)$.
Simplifying we get
$\begin{aligned} & {{\log }_{e}}\left( {{e}^{2y}} \right)={{\log }_{e}}\left( \dfrac{1+x}{1-x} \right) \\\ & \Rightarrow 2y{{\log }_{e}}e=\ln \left( \dfrac{1+x}{1-x} \right) \\\ \end{aligned}$
We have the identity formula of ${{\log }_{x}}x=1$. This gives ${{\log }_{e}}e=1$.
The final form becomes
$\begin{aligned} & 2y=\ln \left( \dfrac{1+x}{1-x} \right) \\\ & \Rightarrow y=\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x} \right) \\\ \end{aligned}$
As we have ${{\tanh }^{-1}}\left( x \right)=y$, we get ${{\tanh }^{-1}}\left( x \right)=y=\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x} \right)$.
Thus proved, ${{\tanh }^{-1}}\left( x \right)=\dfrac{1}{2}\ln \left( \dfrac{1+x}{1-x} \right)$.

Note: Hyperbolic functions are related to the natural exponential function as well the circular sine and cosine functions. They are called “hyperbolic” because the relationship between the sine and cosine functions is the same as a unit hyperbola.