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Math Help - proving that arctanh(x) is differentiable

  1. #1
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    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?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by maximus101 View Post
    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
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  3. #3
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    Quote Originally Posted by chisigma View Post
    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.
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  4. #4
    MHF Contributor chisigma's Avatar
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    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
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  5. #5
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    ok I understand now, thank you
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