# Thread: Differentiation of inverse tanh

1. ## Differentiation of inverse tanh

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!

2. Originally Posted by skeltonjoe
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

$\displaystyle \frac{dy}{dx} = \frac{2}{(1 - x)^2}\cdot \frac{1 - x}{2(1 + x)}$

$\displaystyle =\frac{1}{(1 - x)(1 + x)}$

$\displaystyle = \frac{1}{1 - x^2}$.

3. Originally Posted by skeltonjoe
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!
using log properties

$\displaystyle y = tanh(X) = \frac{1}{2} \ln{\frac{1+x}{1-x}} = \frac{1}{2}(\ln{(1+x)} - \ln{(1-x)})$

so

$\displaystyle y' = \frac{1}{2}\bigg(\frac{1}{1+x} +\frac{1}{1-x}\bigg) = \frac{1}{2}\bigg(\frac{2}{1-x^2}\bigg) = \frac{1}{1-x^2}$

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# prove derivative of tanh^-1 x

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