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 $f_n'(x)$?


2 Answers 2


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

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


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 $c_{2n-1}=0$


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