Exercise on radical ideal and formal derivatives

Question

I need some help for solving the following exercise, because at the moment I'm a little bit lost and don't know where to start.

Given a field $k$ with $\mathrm{char}(k)=0$ and a polynomial $f\in k[x_1,...,x_n]$, where $f=f_{1}^{d_1}f_2^{d_2}\cdots f_m^{d_m}$ with the $f_i$'s being distinct, irreducible, consider the principal ideal $(f)$. I want to show that the radical $\sqrt{(f)}$ equals the principal ideal $(f')$ with $$f'=\frac{f}{\gcd(f,f_{x_1},...,f_{x_n})}$$ where $f_{x_i}$ denotes the formal derivative of $f$ by $x_i$.

A starting point and/or a little hint would be very nice.

Edit: I only know that by definition $\sqrt{(f)}=\{g\in k[x_1,...,x_n]|\exists M\exists h:g^M=hf\}$

Answer 1

I suppose you know that $\sqrt{(f)}=(f_1\cdots f_m)$. Then the question reduces to show $$\gcd(f,f_{x_1},\dots,f_{x_n})=f_1^{d_1-1}\cdots f_m^{d_m-1}.$$ Can you prove this?