Another odd connection between hyperbolics and Phi

Consider tanh'/tanh = 1, the point where tanh = e^x

The solution is:
 (1-tanh^2)/tanh = 1/tahn-tanh = 1 
and we know then that
tanh = 1/Phi.

The angle where that happens is: 1.5 * log(Phi)

But that angle is not one of our Lucas angles, which are integer multiple of Phi.
But we know that:
tanh(x/2) = (cosh(x)-1)/sinh(x), so this drops out of the divisibility of Lucas polynomials, though I have not worked it.

For the trigonometric functions we have:
tan'/tan = 1/tan + tanh  = 1

We get a similar result.  It may seem odd, but circular functions are maximally divergent and we just get the connection between the log,pi and phi; because phi is the most irrational number and will be the maximally divergent.