inverse tanh(X) = 1/2 ln(1+x/1-x) differentiate to show that d/dx = 1/1-x^2
I have been working on this problem for so long, can anyone show me the solution. Thanks!
$\displaystyle \tanh^{-1}{x} = \frac{1}{2}\ln{\frac{1 + x}{1 - x}}$.
So if $\displaystyle y = \tanh^{-1}{x}$
$\displaystyle y = \frac{1}{2}\ln{\frac{1 + x}{1 - x}}$
Let $\displaystyle u = \frac{1 + x}{1 - x}$ so that $\displaystyle y = \frac{1}{2}\ln{u}$.
$\displaystyle \frac{du}{dx} = \frac{(1 - x)\frac{d}{dx}(1 + x) - (1 + x)\frac{d}{dx}(1 - x)}{(1 - x)^2}$ by the Quotient Rule
$\displaystyle = \frac{1 - x + 1 + x}{(1 - x)^2}$
$\displaystyle = \frac{2}{(1 - x)^2}$.
$\displaystyle \frac{dy}{du} = \frac{1}{2}\cdot \frac{1}{u}$
$\displaystyle = \frac{1}{2}\cdot \frac{1}{\frac{1 + x}{1- x}}$
$\displaystyle = \frac{1}{2}\cdot \frac{1 - x}{1 + x}$
$\displaystyle = \frac{1 - x}{2(1 + x)}$.
Therefore
$\displaystyle \frac{dy}{dx} = \frac{2}{(1 - x)^2}\cdot \frac{1 - x}{2(1 + x)}$
$\displaystyle =\frac{1}{(1 - x)(1 + x)}$
$\displaystyle = \frac{1}{1 - x^2}$.