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    Hi, just doing a bit on hyperbolics and a question has asked me to derive the logarithmic form of artanh. I can derive arsinh and arcosh but I'm fairly stumped with this one :/.
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    \displaystyle y=tanh(x)=\frac{e^{2x}-1}{e^{2x}+1} take the bottom fraction over to the otherside. rearrange for e^2x then take logs.

    we then have that x=arctanhy=something else...
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    IIRC, it's easier to derive that the logarithmic forms of arcsin and arccos.

    Let y = tanh x = sinh x / cosh x; so y = \dfrac{e^x-e^{-x}}{e^x+e^{-x}} = \dfrac{e^{2x}-1}{e^{2x}+1}; can you rearrange for x?

    [the reason it's easier is that one doesn't need to solve a quadratic, it can be rearranged linearly.]
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    yay i got it , artanhx = 1/2 ln((x+1)/(1-x))
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    i did this question today, fun times
 
 
 
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Updated: February 13, 2010

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