# Derivative of polynomial all those roots are not simple(multiplicity > 1)

Let $$\operatorname{char}\mathbb K = 0$$. Consider the polynomial $$q(x) := (x - a_1)^{n_1}...(x - a_m)^{n_m} \in \mathbb K[x]$$ and the polynomial $$Q(x) := q(x)(x - a_1)...(x - a_m) \in \mathbb K[x],$$ where $$n_i \in \mathbb N \setminus\{0\}$$ and $$a_i \neq a_j, i \neq j$$. Now consider $$f(x) := \frac{Q^{'}}{q} \in \mathbb K[x].$$ Is it true that all roots of $$f$$ are simple(of multiplicity 1)?

I can show that this is true for $$m \leq 3$$. I'm doing this by directly writing down $$f(x)$$ and calculating it's discriminant.

Any help is welcome

By my reckoning, this is true only when $$\mathbb K$$ is an ordered field. If you can give nonzero numbers $$w_j\in\mathbb K$$ with $$\sum \beta_jw_j=\sum\beta_jw_j^2=0$$ (here $$\beta_j$$ are integers $$\ge 2$$), then you will have an example where it fails. You can achieve this, for example, by taking $$\beta_j=2$$ for all $$j=1,2,3$$ and $$w_j$$ the cube roots of unity in $$\Bbb C$$. [So, I don't believe you have a proof for $$m=3$$.] However, in an ordered field (e.g., $$\Bbb R$$), there can be only the trivial solution of the second equation.
Here's how you get to this. Set $$q(x) = \prod_{i=1}^m (x-a_i)^{\gamma_i-1}$$ (with $$\gamma_i\ge 0$$ integers) and $$Q(x) = \prod_{i=1}^m (x-a_i)^{\gamma_i}$$. You set $$f = Q'/q$$. By logarithmic differentiation (or not), $$\frac{Q'(x)}{Q(x)} = \sum \frac{\gamma_i}{x-a_i},$$ so $$f(x)=\sum \gamma_i \prod_{j\ne i} (x-a_j) = \underbrace{\prod_j (x-a_j)}_{g(x)} \underbrace{\sum \frac{\gamma_i}{x-a_i}}_{h(x)}.$$ Now, suppose $$f(w)=0$$. We want to know if we can also have $$f'(w)=0$$. Note that $$g(w)\ne 0$$, since $$w$$ cannot be one of the $$a_i$$, and so we must have $$h(w)=0$$. By the product rule, $$f'(w)=g(w)h'(w)$$, so $$f'(w)=0 \iff h'(w)=0$$. Note that $$h(w) = \sum\frac{\gamma_i}{w-a_i} \quad\text{and}\quad h'(w)=-\sum\frac{\gamma_i}{(w-a_i)^2}.$$ Since $$\gamma_i>0$$, there can be no solutions of $$h'(w)=0$$ in an ordered field, but, as I pointed out at the beginning, we can certainly have nontrivial solutions to both equations, e.g., in $$\Bbb C$$.