General formulas for derivative of $f_n(x)=\dfrac{ax^n+bx^{n-1}+cx^{n-2}+\cdots}{a'x^n+b'x^{n-1}+c'x^{n-2}+\cdots},\quad a'\neq0$ For the function $f_1(x)=\dfrac{ax+b}{a'x+b'},\quad a'\neq0$ , we have $$f_1'(x)=\dfrac{\begin{vmatrix}{a} && {b} \\ {a'} && {b'}\end{vmatrix}}{(a'x+b')^2}$$
For $f_2(x)=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'},\quad a'\neq0$, we have
$$f_2'(x)=\dfrac{{ \begin{vmatrix}{a} && {b} \\ {a'} && {b'}\end{vmatrix} }x^2+2{ \begin{vmatrix}{a} && {c} \\ {a'} && {c'}\end{vmatrix} }x+{ \begin{vmatrix}{b} && {c} \\ {b'} && {c'}\end{vmatrix} }}{(a'x^2+b'x+c')^2}$$
Can we generalize the formula containing determinants to find $f_n'(x)$?

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