We know about standard partial fraction decomposition that says if $f$ and $g$ are two non-zero polynomials over a field $K$ with $g = \displaystyle\prod_{i=1}^k p_i^{n_i}$ being a product of irreducible polynomials, then we have unique polynomials $h$ and $a_{ij}$ with $\deg a_{ij} < \deg p_i$ such that $\frac{f}{g} = h + \displaystyle\sum_{i=1}^k\displaystyle\sum_{j=1}^{n_i} \frac{a_{ij}}{p_i^j}$ with the convention that $\deg f < \deg g$ means that $h = 0$.
I was curious about a special case of when $f = g'$, namely what the partial fraction decomposition would look like for $\frac{g'(x)}{g(x)}$ where $g(x)$ has zeroes $a_1,\dots,a_k$ with multiplicities $m_1,\dots,m_k$.
My thinking is that it should decompose perfectly into $\sum_{i=1}^k \frac{m_i}{x-a_i}$ based on the statement about the original partial fraction decomposition, however I am unsure how this may arise algebraically with a full proof. How might one prove the partial fraction decomposition for this special case?