proving that arctanh(x) is differentiable

• Feb 18th 2011, 07:04 AM
maximus101
proving that arctanh(x) is differentiable
I know that $\displaystyle x$ $\displaystyle \mapsto$ $\displaystyle logx$

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

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

is also differentiable?
• Feb 18th 2011, 07:12 AM
chisigma
Quote:

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

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

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

is also differentiable?

Is [simply]...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 18th 2011, 07:31 AM
maximus101
Quote:

Originally Posted by chisigma
Is [simply]...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Hey, thank you for the answer, but could you please also tell me how differentiability is possible, i.e prove that $\displaystyle ln(1+x)$ and $\displaystyle ln(1-x)$ is differentiable then I can use the sum, quotient rules as you have done.
• Feb 18th 2011, 09:22 AM
chisigma
As You have told is $\displaystyle \displaystyle \frac{d}{dx} \ln x = \frac{1}{x}$ , so that if You instead of $\displaystyle x$ write $\displaystyle 1+x$ is $\displaystyle \displaystyle \frac{d}{dx} \ln (1+x) = \frac{1}{1+x}$, and if You write $\displaystyle 1-x$ is $\displaystyle \displaystyle \frac{d}{dx} \ln (1-x) = -\frac{1}{1-x}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 20th 2011, 04:24 AM
maximus101
ok I understand now, thank you :)