Why is the golden ratio lurking in $(d/dx)\arctan\left( x + \frac{1}{x} \right)$
$$
= \frac{\left(\frac{1+\sqrt{5}}{2}\right)}{x^2 + \left(\frac{1+\sqrt{5}}{2}\right)^2} + \frac{\left(\frac{1-\sqrt{5}}{2}\right)}{x^2 + \left(\frac{1-\sqrt{5}}{2}\right)^2}\text{ ?}
$$
Is this merely an instance of its (unbeknownst to me) lurking everywhere, or is something special about this particular arctangent of a sum?
(An arctangent of a sum seems like a bit of a freak, though.)
(This was inspired by a related question that someone posted to http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics.)
Best Answer
"Welcome to $K_1( \mathbb{C}(t))$!"
The identity instantiates the fact that if $f(z)$ is a rational function, the complex, multivalued $\log(f(z))$ is a sum of logarithms of the linear factors of $f(z)$. This fact can be made single-valued (by differentiating the identity) and real (by taking the real or, in this case, imaginary part of the formula, i.e., symmetrizing under Gal(C/R) which replaces logarithm with arctangent).
Specializing the fact to $f(x)=g(ix)$, where $g(x) = x^2 - x - 1$ and $x$ is real, produces the identity with the golden ratio. The imaginary part of $\log(f)$ is $(1/2i)\log(g(ix)/g(-ix))$, which can be expanded as a sum over the roots and differentiated.
(Remember also that $\arctan t = \arg (q+itq)$ for real $t$ and $q$, so that $\log f(x)$ can be evaluated without factorization, by computing real and imaginary parts of $g(ix)$. Equating the two expressions for the imaginary part of ($d\log(f)$) gives the formula in the question.)