I am trying to find a polynomial for the $n$-th derivative of a quotient, AKA find $$D_n(x)=\left(\frac{d}{dx}\right)^n\frac{f(x)}{g(x)}$$ I started by noting that $$D_{n}(x)=\frac{d}{dx}D_{n-1}(x)$$ And then I noted that $$g(x)D_0(x)=f(x)$$ So applying $\left(\frac{d}{dx}\right)^n$ on both sides, $$\sum_{k=0}^{n}{n\choose k}D_{n-k}(x)g^{(k)}(x)=f^{(n)}(x)$$ Hence a recurrence relation: $$D_n(x)=\frac{f^{(n)}(x)}{g(x)}-\frac1{g(x)}\sum_{k=1}^{n}{n\choose k}D_{n-k}(x)g^{(k)}(x)$$ But I do not know what to do with this.
Another avenue I considered is finding the polynomial $P_n(x)$ such that $$D_n(x)=\frac{P_n(x)}{g(x)^{2^n}}$$ I thought of this because $$D_1(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$$ So $$P_1(x)=f'(x)g(x)-f(x)g'(x)$$ Which gives $$D_2(x)=\frac{P_1'(x)g(x)-P_1(x)[g(x)^2]'}{g(x)^4}$$ Hence $$P_2(x)=P_1'(x)g(x)-P_1(x)[g(x)^2]'$$ And from the same sort of method, I was able to show that $$P_n(x)=g(x)^{2^{n-1}}P_{n-1}'(x)-2^{n-1}g(x)^{2^{n-1}-1}g'(x)P_{n-1}(x)$$ Honestly I think this method is more productive, but I still don't know where to go from here. Could I have some help? Thanks.
Apparently $$\left(\frac{d}{dx}\right)^n f[g(x)]=n!\sum_{n=\sum_{k=1}^{n}km_k\\ (m_1,...,m_n)\in\Bbb N_0^n}f^{(m_1+...+m_n)}[g(x)]\cdot\prod_{j=1}^{n}\frac1{m_j!}\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j}$$ Which is Faà di Bruno's formula. Which is, in itself, rather amazing. Another form of this is given as $$\left(\frac{d}{dx}\right)^n f[g(x)]=\sum_{k=0}^{n}f^{(k)}[g(x)]\cdot B_{n,k}\left[g'(x),g''(x),..,g^{(n-k+1)}(x)\right]\qquad\qquad (1)$$ Where $B_{n,k}(x_1,x_2,...,x_{n-k+1})$ is a Bell Polynomial.
We use this by considering $f/g$ as the product $f\cdot\frac1g$ and using the general Leibniz rule as follows: $$D_n(x)=\sum_{k=0}^{n}f^{(n-k)}(x)\left[\left(\frac{d}{dx}\right)^k \frac1{g(x)}\right]$$ Anyway, we use $f(x)=1/x$ in $(1)$: $$\left(\frac{d}{dx}\right)^k 1/g(x)=\sum_{k=0}^{n} B_{n,k}\left[g'(x),g''(x),..,g^{(n-k+1)}(x)\right]\cdot \left[\left(\frac{d}{dz}\right)^k \frac1z \right]_{z=g(x)}$$ $$\left(\frac{d}{dx}\right)^k 1/g(x)=\sum_{k=0}^{n} \frac{(-1)^k k!}{g(x)^{k+1}}B_{n,k}\left[g'(x),g''(x),..,g^{(n-k+1)}(x)\right]$$ So at long last, $$D_n(x)=\sum_{k=0}^{n}f^{(n-k)}(x)\sum_{r=0}^{k} \frac{(-1)^r r!}{g(x)^{r+1}}B_{k,r}\left[g'(x),g''(x),..,g^{(k-r+1)}(x)\right]$$ Which is the most amazing derivative rule I've ever seen. Thanks to @nathan.j.mcdougall for commenting the original link.