Derivatives of the nth cyclotomic polynomial Are there any useful properties of the $k$th derivative of the $n$th cyclotomic polynomial? 
In particular, what would the value of this be at $1$ and $0$, or any properties of the value of the $k$th derivative at $0$ or $1$?
 A: The derivative $\Phi_n'(t)$ is useful not only for $t=0$ or $t=1$, but for $t=\zeta_{p^r}$, a $p^r$-th primitive root of unity. To show that of all prime ideals in $\mathbb{Z}$, only $P=(p)$ is ramified in the cyclotomic field $K=\mathbb{Q}(\zeta_{p^r})$, one needs that the discriminant is given by
$$
D(1,\zeta_{p^r},\ldots ,\zeta_{p^r}^{\phi(p^r)})=\pm N_{K/\mathbb{Q}}(\Phi_n'(\zeta_{p^r}))=\pm p^{rp^r-rp^{r-1}-p^{r-1}}.
$$
A: Here are some results for the first derivative at $0$ and at $1$. Perhaps you can exploit the technique to get more of these. I suspect it'll get messy.

Theorem: We have $\Phi_1'(0) = 1$ and $$\Phi_n'(0) = -\mu(n)$$ for $n>1$, where $\mu$ is the Möbius function.

Proof: We have
$$X^n-1 = \prod_{d \mid n} \Phi_d(X).$$
Taking logarithmic derivatives on both sides, we have
$$\frac{nX^{n-1}}{X^n-1} = \sum_{d\mid n} \frac{\Phi'_d(X)}{\Phi_d(X)}.$$
None of these terms have poles at $X=0$, so we can just put $X=0$ on both sides to get (for $n>1$):
$$0 = \sum_{d\mid n}\frac{\Phi'_d(0)}{\Phi_d(0)}.$$
Now we have $\Phi_d(0) = -1$ for $d=1$ and $\Phi_d(0) = 1$ for $d>1$. Thus we can rewrite this as
$$0 = 1 + \sum_{d \mid n, d>1}-\Phi_d'(0).$$ By Möbius inversion we get the result.

Theorem: We have $\Phi_n'(1)=1$, and $$\Phi_n'(1) = e^{\Lambda(n)} \varphi(n)/2$$ for $n>1$, where $\Lambda$ is the Von Mangoldt function
  and $\varphi$ is Euler's function.

The proof is similar. In the equation
$$\frac{nX^{n-1}}{X^n-1} = \sum_{d\mid n} \frac{\Phi'_d(X)}{\Phi_d(X)},$$
we subtract $\Phi_1'(X)/\Phi_1(X) =1/(X-1)$ from both sides, because we want to get rid of the pole. Then we get, after cancellation,
$$\frac{(n-1)X^{n-2} + (n-2)X^{n-3} + \dots + 2X + 1}{X^{n-1} + \dots + X + 1} = \sum_{d\mid n, d >1} \frac{\Phi'_d(X)}{\Phi_d(X)}.$$
Evaluating both sides at $X=1$, we get
$$\frac{n-1}{2} = \sum_{d\mid n, d >1} \frac{\Phi'_d(1)}{\Phi_d(1)}.$$ Once again, by Möbius inversion, by the fact that $\Phi_d(1) = e^{\Lambda(d)}$ for $d>1$ (exercise), and by the fact that $\sum_{d \mid n} \mu(d) d = \varphi(n)$, we get the result.
