# Question #71b42

Oct 12, 2017

The solution is $y = \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right)$

#### Explanation:

By definition

$\tanh x = \sinh \frac{x}{\cosh} x$

$\cosh x = \frac{{e}^{x} + {e}^{-} x}{2}$

$\sinh x = \frac{{e}^{x} - {e}^{-} x}{2}$

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

Then,

$x = \tanh y$

$= \sinh \frac{y}{\cosh} y$

$= \frac{\frac{{e}^{y} - {e}^{-} y}{2}}{\frac{{e}^{y} + {e}^{-} y}{2}}$

$= \frac{\left({e}^{y} - {e}^{-} y\right)}{\left({e}^{y} + {e}^{-} y\right)}$

$= \frac{\left({e}^{y} - \frac{1}{e} ^ y\right)}{\left({e}^{y} + \frac{1}{e} ^ y\right)}$

$= \frac{{e}^{2 y} - 1}{{e}^{2 y} + 1}$

So,

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

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

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

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

Taking natural logs

$\ln \left({e}^{2 y}\right) = \ln \left(\frac{1 + x}{1 - x}\right)$

$2 y = \ln \left(\frac{1 + x}{1 - x}\right)$

$y = \frac{1}{2} \ln \left(\frac{1 + x}{1 - x}\right)$

Conditions .

$\forall x < | x |$