Let $n \geq 2$ and $f:\mathbb{R} \to \mathbb{R}, \: f(x)=(x-x_1)(x-x_2)\dots(x-x_n)$ where $x_1,\dots, x_n$ are distinct real numbers. The matrix $A=(a_{ij})_{1 \leq i,j \leq n}$ is defined as follows: $$a_{ij}=\begin{cases} \dfrac{f'(x_i)-f'(x_j)}{x_i-x_j}, & \text{if } i \neq j \\[6px] f''(x_i), & \text{if } i = j \end{cases}$$ Prove that $\det A = 0$
This is related to this problem, so I know that $$\frac{1}{f(x)}=\sum_{k=1}^n \frac{1}{f'(x_k)(x-x_k)}, \quad \forall x \neq x_i$$ I tried to write the terms of $A$, but getting the second derivative of $f$ in terms of those $x_i$ is pretty complicated. So I tried to simplify it by writing $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$$ and so $f''(x)=n(n-1)x^{n-2}+(n-1)(n-2)x^{n-3}+\dots+2\cdot 1\cdot x^0$. However, I am not sure this will lead to something helpful.