You can easily extend this, but for $n\geq 3$ you will end up with more than one term per monomial:

For two functions $f$, $g$ rewrite the quotient rule using a determinant

$$\frac{d}{dx} \frac{f}{g} = \frac{\frac{df}{dx}g-f \frac{dg}{dx}}{g^2} = \frac{\begin{vmatrix} \frac{df}{dx} & f \\ \frac{dg}{dx} & g \end{vmatrix}}{g^2}$$

Now assume that $f(x) := a_nx^n+\dots + a_0$, $g(x) :=b_n x^n+ \dots + b_0$ are polyonomials. Then $\frac{df}{dx}$ and $\frac{dg}{dx}$ can be calculated explicitly and you can use the multilinearity of the determinant to split it by monomials:

$$\frac{d}{dx} \frac{f}{g} = \frac{\begin{vmatrix} \sum_{k=0}^n a_k k x^{k-1} & \sum_{j=0}^n a_j x^j \\ \sum_{k=0}^n b_k k x^{k-1} & \sum_{j=0}^n b_j x^j \end{vmatrix}}{g^2} = \frac{\sum_{k=0}^n \sum_{j=0}^n k\begin{vmatrix} a_k & a_j \\ b_k & b_j \end{vmatrix} x^{k+j-1} }{g^2}$$

Now in the last sum for $k=j$ the determinant vanishes and if $k\neq j$ the same determinant occurs again with flipped sign if their roles are reversed. So you can only count the cases $j<k$ and get

$$\frac{d}{dx} \frac{f}{g} = \frac{\sum_{k=0}^n \sum_{j=0}^{k-1} (k-j)\begin{vmatrix} a_k & a_j \\ b_k & b_j \end{vmatrix} x^{k+j-1} }{g^2} $$