I'm currently working on the following exercise (Lang, IV.11(c)):
Let $D$ be the standard derivative in the polynomial ring $k[X]$ over a field $k$. Let $R(X)=c\prod_{j}(X-\alpha_{j})^{m_{j}}$ with $\alpha_{j}\in k$, $c\in k$, and $m_{j}\in\mathbb{Z}$ (so $R$ is a rational function). Show that $$\frac{R'(X)}{R(X)}=\sum_{j}\frac{m_{j}}{X-\alpha_{j}}.$$
My thoughts: I know that if $A$ is a commutative ring, then the derivative is a map $D:A[X]\to A[X]$ defined as expected: $$Df(X)=f'(X)=a_{1}+2a_{2}X+\cdots+na_{n}X^{n-1}$$ for $f(X)=a_{0}+a_{1}X+a_{2}X^{2}+\cdots+a_{n}X^{n}$. I would think that I can apply this in some sort of pairwise product rule fashion to take the derivative of $R(X)$. Or perhaps I could multiply all of the products out? However, I'm not sure how exactly to do this since we just have a general expression for a product of linear factors.
My questions: Do either of my methods mentioned above make sense? Also, the indexing set for $j$ must be finite, right? (If not, then we would be dealing with the ring of power series $k[[X]]$, not the ring of polynomials $k[X]$.)
Thanks in advance for any help.
The key is to generalize the Leibniz rule $(fg)'=f'g+fg'$ to multiple factors: we have $$D\left(\prod_j f_j\right)=\sum_if_i'\prod_{j\neq i}f_j$$ Thus, for $R(X)=c\prod_{j}(X-\alpha_{j})^{m_{j}}$ we have \begin{align} R'(X) &=cD\left(\prod_{j}(X-\alpha_{j})^{m_{j}}\right)\\ &=c\sum_{i}D((X-\alpha_{i})^{m_i})\prod_{j\neq i}(X-\alpha_{j})^{m_{j}}\\ &=c\sum_{i}m_i(X-\alpha_{i})^{m_i-1}\prod_{j\neq i}(X-\alpha_{j})^{m_{j}} \end{align} hence \begin{align} \frac{R'(X)}{R(X)} &=\frac{c\sum_{i}m_i(X-\alpha_{i})^{m_i-1}\prod_{j\neq i}(X-\alpha_{j})^{m_{j}}}{c\left(\prod_{j}(X-\alpha_{j})^{m_{j}}\right)}\\ &=\sum_{i}m_i(X-\alpha_{i})^{m_i-1}\frac{\prod_{j\neq i}(X-\alpha_{j})^{m_{j}}}{\prod_{j}(X-\alpha_{j})^{m_{j}}}\\ &=\sum_{i}m_i(X-\alpha_{i})^{m_i-1}\frac{1}{(X-\alpha_i)^{m_i}}\\ &=\sum_{i}\frac{m_i}{X-\alpha_i}\\ \end{align}