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Do arccoth(x) and arctanh(x) have the same derivatives?

  1. May 15, 2005 #1
    Hello, I am wondering if arccoth(x) and arctanh(x) have the same derivatives? I ask this because I got:

    [tex]
    \frac{d}{{dx}}arc\coth (x) = \frac{1}{{1 - x^2 }} = \frac{d}{{dx}}\arctan h(x)
    [/tex]

    I don't think their derivatives should be the same. Does it have anything to do with domain restrictions? Any help appreciated.
     
  2. jcsd
  3. May 15, 2005 #2

    arildno

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    Well, you have:
    [tex]artanh(x)=\frac{1}{2}ln(\frac{1+x}{1-x}), |x|<1, arcoth(x)=\frac{1}{2}ln(\frac{1+x}{x-1}), |x|>1[/tex]

    Note that, in general, the derivatives of ln(x) and ln(-x) with respect to "x" is the same; but their domains are disjoint.
     
    Last edited: May 15, 2005
  4. May 15, 2005 #3
    Thanks for the help arildno.
     
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