Discriminant of $f(x^n)$ I read about this post: Discriminant of $f(x^n)$ for $f$ a quadratic about $\Delta f(x^n)=x^{2n}-bx^n+c$ if $f(x) = x^2-bx+c$ is a quadratic. In particular, $\Delta f(x^n) = n^{2n}c^{n-1}\Delta(f(x))$.
Is there any relation like this for general $f$?
 A: There is a relation like this for general $f$. To avoid leading term annoyances, every polynomial referenced in this answer will be monic -- if your $f$ is not monic, you can simply replace $f$ with $f$ divided by its leading coefficient and figure out the appropriate normalization to apply at the end.
For polynomials $f_1,\dots,f_n$,
$$\Delta(f_1\cdots f_n)=\prod_{i=1}^n\Delta(f_i)\prod_{1\leq i<j\leq n}\operatorname{Res}(f_i,f_j)^2,$$
where
$$\operatorname{Res}(f,g)=\prod_{f(\alpha)=0}g(\alpha)$$
is the resultant and the product is over roots $\alpha$ of $f$ counted with multiplicity. (This result comes directly from the definition of the discriminant as products of differences of roots.) One may compute, for $\beta,\gamma\in\mathbb C$,
$$\Delta(x^n-\beta)=\beta^{n-1}\Delta(x^n-1)=\beta^{n-1}(-1)^{n(n-1)/2+1}n^n$$
and
$$\operatorname{Res}(x^n-\beta,x^n-\gamma)=\prod_{x^n=\beta}(x^n-\gamma)=(\beta-\gamma)^n.$$
This means that, if $S$ is the multiset of roots of a polynomial $f$ (counted with multiplicity), then
\begin{align*}
\Delta(f(x^n))
&=\Delta\left(\prod_{\beta\in S}(x^n-\beta)\right)\\
&=\prod_{\beta\in S}\Delta(x^n-\beta)\prod_{\{\beta,\gamma\}\subset S}\operatorname{Res}(x^n-\beta,x^n-\gamma)^2\\
&=\prod_{\beta\in S}\left(n^n(-1)^{n(n-1)/2+1}\beta^{n-1}\right)\prod_{\{\beta,\gamma\}\subset S}(\beta-\gamma)^{2n}.
\end{align*}
The first term multiplies to
\begin{align*}
n^{n\deg f}(-1)^{\left(\frac{n(n-1)}2+1\right)\deg f}\left(\prod_{\beta\in S}\beta\right)^{n-1}
&=n^{n\deg f}(-1)^{\left(\frac{n(n-1)}2+1\right)\deg f}\left((-1)^{\deg f}f(0)\right)^{n-1}\\
&=n^{n\deg f}(-1)^{\left(\frac{n(n+1)}2\right)\deg f}f(0)^{n-1}.
\end{align*}
The second term is simply $\Delta(f)^n$. So,
$$\Delta\left(f(x^n)\right)=\left((-1)^{n(n+1)/2}n^n\right)^{\deg f}f(0)^{n-1}\Delta(f)^n.$$
