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 $f_n'(x)$?

If $f_n=\frac{P_n(x)}{Q_n(x)}$ and $P_n=\sum a_kx^k$,
$Q_n=\sum b_kx^k$, then
$$f'_n=\frac{\begin{vmatrix}{P'} && {Q'} \\ {P} && {Q}\end{vmatrix} }{Q^2}$$
Breaking the determinant on the numerator gives $$\sum_{j\geq 0} \left(\sum_{k+r=j+1} k\begin{vmatrix}{a_{k}} && {b_k} \\ {a_{j+1-k}} && {b_{j+1-k}}\end{vmatrix} \right)x^j=\sum_{j=0}^{2n-1}c_jx^j$$
Now, $k,r \leq n$ implies $ n\geq k,r \geq (j+1)-n ; k>0$.
Also, we can further simplify $c_j$ as $$c_j=\sum_{k=j+1-n}^{\lfloor{\frac{j+1}{2}}\rfloor} k\begin{vmatrix}{a_k} && {b_k} \\ {a_{j+1-k}} && {a_{j+1-k}}\end{vmatrix}+ (j+1-k)\begin{vmatrix}{a_{j+1-k}} && {b_{j+1-k}} \\ {a_k} && {b_k}\end{vmatrix}$$
Hence, $$c_j=\sum_{k=j+1-n}^{\lfloor{\frac{j+1}{2}}\rfloor} (j+1-2k)\begin{vmatrix}{a_{j+1-k}} && {b_{j+1-k}} \\ {a_{k}} && {b_{k}}\end{vmatrix}$$
This gives $c_{2n-1}=0$
