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

1. May 15, 2005

### Benny

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

$$\frac{d}{{dx}}arc\coth (x) = \frac{1}{{1 - x^2 }} = \frac{d}{{dx}}\arctan h(x)$$

I don't think their derivatives should be the same. Does it have anything to do with domain restrictions? Any help appreciated.

2. May 15, 2005

### arildno

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

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
3. May 15, 2005

### Benny

Thanks for the help arildno.