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

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Arccoth(x) and arctanh(x) do not have the same derivatives due to their differing domains. The derivative of arccoth(x) is given by \(\frac{d}{{dx}}arc\coth (x) = \frac{1}{{1 - x^2 }}\), while the derivative of arctanh(x) is defined only for \(|x| < 1\). The functions themselves are defined as arctanh(x) = \(\frac{1}{2}ln(\frac{1+x}{1-x})\) for \(|x| < 1\) and arccoth(x) = \(\frac{1}{2}ln(\frac{1+x}{x-1})\) for \(|x| > 1\). The disjoint nature of their domains is crucial in understanding their derivatives.

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Hello, I am wondering if arccoth(x) and arctanh(x) have the same derivatives? I ask this because I got:

<br /> \frac{d}{{dx}}arc\coth (x) = \frac{1}{{1 - x^2 }} = \frac{d}{{dx}}\arctan h(x)<br />

I don't think their derivatives should be the same. Does it have anything to do with domain restrictions? Any help appreciated.
 
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Well, you have:
artanh(x)=\frac{1}{2}ln(\frac{1+x}{1-x}), |x|&lt;1, arcoth(x)=\frac{1}{2}ln(\frac{1+x}{x-1}), |x|&gt;1

Note that, in general, the derivatives of ln(x) and ln(-x) with respect to "x" is the same; but their domains are disjoint.
 
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Thanks for the help arildno.
 

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