<s>[INST] Why does the technique to find derivative of $y=\frac{ax^2+bx+c}{a'x^2+b'x+c'}$ quickly, works? I just watched a video about a very good technique to evaluate derivative of $ y=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'} ,\quad a'\neq0$ quickly:
$$\large y'=\large\dfrac{{ \begin{vmatrix}{a} && {b} \\ {a'} && {b'}\end{vmatrix} }x^2+\color{red}2{ \begin{vmatrix}{a} && {c} \\ {a'} && {c'}\end{vmatrix} }x+{ \begin{vmatrix}{b} && {c} \\ {b'} && {c'}\end{vmatrix} }}{(a'x^2+b'x+c')^2}$$
I wonder is it possible to prove this is true? Of course we can calculate derivative of $y$ and expanding the terms and compare it with the formula that I mentioned. But is it possible to check this is true intuitively Or without expanding? ( it is obvious that denominator should be $(a'x^2+b'x+c')^2$  )
 [/INST] Let $f(x)=ax^2+bx+c$ and $g(x)=a'x^2+b'x+c'$. Then
\begin{aligned}
\frac{d(f/g)}{dx}
&=\frac{f'g-g'f}{g^2}\\
&=\frac{1}{g^2}\left|\begin{matrix}1&0&0\\ 0&f'&f\\ 0&g'&g\end{matrix}\right|\\
&=\frac{1}{g^2}\left|\begin{matrix}1&0&0\\ a&f'&f\\ a'&g'&g\end{matrix}\right|\quad(R_2=R_2+aR_1;\ R_3=R_3+a'R_1)\\
&=\frac{1}{g^2}\left|\begin{matrix}1&-2x&0\\ a&b&f\\ a'&b'&g\end{matrix}\right|\quad(C_2=C_2-2xC_1)\\
&=\frac{1}{g^2}\left|\begin{matrix}1&-2x&x^2\\ a&b&c\\ a'&b'&c'\end{matrix}\right|\quad(C_3=C_3-x^2C_1-xC_2)\\
&=\frac{1}{g^2}\left(\left|\begin{matrix}b&c\\ b'&c'\end{matrix}\right|+2x\left|\begin{matrix}a&c\\ a'&c'\end{matrix}\right|+x^2\left|\begin{matrix}a&b\\ a'&b'\end{matrix}\right|\right).
\end{aligned}
</s>