Derivatives of Polynomia and Other Factorizable Functions

Let the function $p$ be a polynomium of degree $n$, with $n$ roots $\{x_i\}\in \mathbb C$. We may then write $p$ as \begin{align} p(x) &= a\cdot \prod_{i=1}^n \left(x - x_i\right)\,, \end{align} where $a$ is a constant. We wish to obtain an expression for the derivative of $p$. Consider the derivative of $\ln(p(x))$: \begin{align} \frac{d}{dx}\ln(p(x)) &= \frac{d}{dx}\ln\left(a\cdot \prod_{i=1}^n \left(x - x_i\right)\right)\\ &= \frac{d}{dx}\left(\ln(a) + \sum_{i=1}^{n}\ln\left(x-x_i\right)\right)\\ &= \sum_{i=1}^{n} \frac{d}{dx} \ln\left(x-x_i\right)\\ &=\sum_{i=1}^{n} \frac{1}{x-x_i}\,. \end{align} On the other hand, we can take the logarithmic derivative of $p$ directly \begin{align} \frac{d}{dx}\ln(p(x)) &= \frac{p'(x)}{p(x)}\,. \end{align} It follows that \begin{align} \frac{p'(x)}{p(x)} &= \sum_{i=1}^{n} \frac{1}{x-x_i} \end{align} or \begin{align} p'(x) &= p(x)\cdot \sum_{i=1}^{n} \frac{1}{x-x_i}\\ &= a\cdot \sum_{i=1}^{n}\frac{\prod_{j=1}^n \left(x - x_j\right)}{x-x_i}\\ &= a\cdot \sum_{i=1}^{n}\prod_{j\neq i}^n (x-x_j)\,. \end{align} An unnecessarily tideous way of taking the derivative of a polynomium is thus to take the sum of the polynomia obtained when factoring out each root in turn.
The method is straightforwardly generalized to any factorizable function, i.e., \begin{align} f(x) &= \prod_{i=1}^n g_i(x) \end{align} has the derivative \begin{align} f'(x) &= f(x)\cdot \sum_{i=1}^n \frac{g_i'(x)}{g_i(x)}\,. \end{align} yielding a generalized product rule for differentiation