Consider the graph of $y=x^n$ with $n$ odd, and draw a tangent to its negative arc that crosses the graph at $(1,1)$. Equating the slopes gives us $\frac{x^n-1}{x-1}=nx^{n-1}$ at the point of tangency. Since $$nx^{n-1}-\frac{x^n-1}{x-1}=(x-1)\left(\frac{x^n-1}{x-1}\right)'$$ this point has to be the only real (negative) root of $$\left(\frac{x^n-1}{x-1}\right)'=(n-1)x^{n-2}+(n-2)x^{n-3}+\dots+2x+1=0.$$ When $n=p$ an odd prime this is just the derivative of the $p$-th cyclotomic polynomial $\Phi_p'$ (not sure if $\frac{x^n-1}{x-1}$ has a special name in general). For $n=3$, $x_3=-\frac12$, and for $n=5$ one can express $x_5$ by Cardano's formula, but it is not very illuminating.
Since cyclotomic polynomials are a classical topic I thought it would be easy to find information on the roots of their derivatives. But my searches in the usual places (MathSciNet and Google Scholar) came up short, only the values of $\Phi_p'$ at $0$ and $1$ are typically discussed. I am mostly interested in some good asymptotics for the negative root $x_p$, but information on complex roots is welcome too, if there is any. From geometry, $x_n$ is obviously monotone decreasing, so the asymptotics is the same for all large odd $n$, and, since $y=x^n$ approaches a vertical step, $x_n\to-1$. But I'd like something more precise, an expansion ideally. Same for $y_n=x_n^n\to0$. I'll appreciate even a hint in the right direction, the typical asymptotic methods for roots seem to have the parameter enter the equation differently.
