Thread: proving that arctanh(x) is differentiable

1. proving that arctanh(x) is differentiable

I know that $x$ $\mapsto$ $logx$

is differentiable on (0, $\infty$) with derivative $1/x$

but how do I use this to prove that $arctanhx$= $(1/2)$ $log$ $(1+x)/(1-x)$= $(1/2)$ $log(1+x)$ - $(1/2)$ $log(1-x)$

is also differentiable?

2. Originally Posted by maximus101
I know that $x$ $\mapsto$ $logx$

is differentiable on (0, $\infty$) with derivative $1/x$

but how do I use this to prove that $arctanhx$= $(1/2)$ $log$ $(1+x)/(1-x)$= $(1/2)$ $log(1+x)$ - $(1/2)$ $log(1-x)$

is also differentiable?
Is [simply]...

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

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
Is [simply]...

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

Kind regards

$\chi$ $\sigma$
Hey, thank you for the answer, but could you please also tell me how differentiability is possible, i.e prove that $ln(1+x)$ and $ln(1-x)$ is differentiable then I can use the sum, quotient rules as you have done.

4. As You have told is $\displaystyle \frac{d}{dx} \ln x = \frac{1}{x}$ , so that if You instead of $x$ write $1+x$ is $\displaystyle \frac{d}{dx} \ln (1+x) = \frac{1}{1+x}$, and if You write $1-x$ is $\displaystyle \frac{d}{dx} \ln (1-x) = -\frac{1}{1-x}$...

Kind regards

$\chi$ $\sigma$

5. ok I understand now, thank you