polynomial for $\left(\frac{d}{dx}\right)^n \frac{f(x)}{g(x)}$ 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.
 A: 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.
