Looking for a nonrecursive formula for the general derivatives of the quotient of functions I want to prove that the $k$-th derivative $h^{(k)}(x)$ of the function $h(x)=\frac{1}{1+x^2}$ is zero at $x=0$ for all integer values $k>0$.
My only idea was to go the stubborn way applying iteratively the elementary formula for the derivative of a quotient of functions. Alas I didnt find a general formula similar to the Leibniz formula for the derivatives of a product of functions neither in Wikipedia nor else in the web so far.
This puzzles me. It wouldnt surprise me if a non-recursive closed expression (using iterated binomial coefficient sums) would be existing.
I tackled the problem so far in using first the Leibniz formula on $h(x)=f(x)\frac{1}{g(x)}$ in the way
$$h^{(k)}(x)=\sum_{r=0}^k f^{(k)}(x)\left( \frac{1}{g(x)}\right)^{(k-r)}$$
So I am in front of the problem calculating $\left( \frac{1}{g(x)}\right)^{(s)}$ which in the first step tackled by decreasing iteratively by $1$
$$\left( \frac{1}{g(x)}\right)^{(s)}=\left( -\frac{g^{(1)}(x)}{g^2(x)}\right)^{(s-1)}$$
If one applies to that quotient iteratively the Leibniz product rule
the next calculation problem comes up for
$$\left( \frac{1}{g^2(x)}\right)^{(t)}=\left( -\frac{(g^2)^{(1)}(x)}{g^4(x)}\right)^{(t-1)}$$
So I arrive at the problem calculating
$$\left( -\frac {g^{{2^{m-1}}^{(1)}}(x)} {g^{2^m}(x)} \right)^{(1)}$$
Then I tried to go on by de-exponentiating and getting the square
$$(g^{2^{m-1}})^{(1)}(x)=\left((g^{2^{m-2}}(x))^2\right)^{(1)}(x)=2g^{2^{m-2}}(x)g^{2^{m-2}})^{(1)}(x)$$
This leads me iteratively to the ( surprising/erroneous(?) ) result
$$\left( -\frac {g^{{2^{m-1}}^{(1)}}(x)} {g^{2^m}(x)} \right)^{(1)}=2^{m-1}\frac{g^{\prime}}{g}$$
where the exponent at the $2$ is in doubt.
Now I am overwhelmed at putting this all together and especially simplifying the nested iterative Leibniz sums with the binomials.

In order to answer OPs second part of the question, we provide an expression of the  $n$-th derivative of $\frac{f}{g}$  in terms of  derivatives  of $f$  and $g$. 

Let $D_x$ represent differentiation with respect to $x$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$.  The following holds true for $n$ times differentiable functions $f$ and $g$
  \begin{align*}
D_x^n\left(\frac{f}{g}\right)=\sum_{k=0}^n\sum_{j=0}^{k} (-1)^j\binom{n}{k}\binom{k+1}{j+1}\frac{1}{g^{j+1}}
D_x^{n-k}\left(f\right) D_x^{k}\left( g^j\right)
\end{align*}

The following formula for the $n$-th derivative of the composite of two functions is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:

Hoppe Form of Generalized Chain Rule
Let $D_z$ represent differentiation with respect to $z$ and $z=z(x)$. Hence $D^n_x f(z)$ is the $n$-th derivative of $f$ with respect to $x$.  The following is valid
  \begin{align*}
D_x^n f(z)=\sum_{k=0}^nD_z^kf(z)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}z^{k-j}D_x^nz^j
\end{align*}

We consider $z=g$
\begin{align*}
f(g)=\frac{1}{g}\qquad\qquad  g=g(x)
\end{align*}
and obtain
\begin{align*}
D_x^n\left(\frac{1}{g}\right)=\sum_{k=0}^nD_g^k\left(\frac{1}{g}\right)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}g^{k-j}D_x^ng^j\tag{1}
\end{align*}
Since
\begin{align*}
D_g^k\left(\frac{1}{g}\right)=(-1)^kk!\frac{1}{g^{k+1}}
\end{align*}
We obtain from (1)
\begin{align*}
D_x^n\left(\frac{1}{g}\right)&=\sum_{k=0}^n\sum_{j=0}^k(-1)^j\binom{k}{j}\frac{1}{g^{j+1}}D_x^ng^j\\
&=\sum_{j=0}^n(-1)^j\frac{1}{g^{j+1}}D_x^ng^j\sum_{k=j}^n\binom{k}{j}\tag{2}\\
\end{align*}
Since the following identity holds,
\begin{align*}
\sum_{k=j}^n\binom{k}{j}=\binom{n+1}{j+1}
\end{align*}

we finally obtain from (2)
  \begin{align*}
D_x^n\left(\frac{1}{g}\right)=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{1}{g^{j+1}}D_x^ng^j\tag{3}\\
\end{align*}
  This identity is also stated as (3.63) in H.W. Goulds book, Vol. I.
Next  we consider the $n$-th derivative of $f$ with $\frac{1}{g}$ and use  the Leibniz formula. We derive
  \begin{align*}
D_x^n\left(\frac{f}{g}\right)&=\sum_{k=0}^n \binom{n}{k}D_x^k\left(f\right)D_x^{n-k}\left(\frac{1}{g}\right)\\
&=\sum_{k=0}^n\sum_{j=0}^{n-k} (-1)^j\binom{n}{k}\binom{n-k+1}{j+1}\frac{1}{g^{j+1}}
D_x^k\left(f\right)D_x^{n-k}\left(g^j\right)\tag{4}\\
&=\sum_{k=0}^n\sum_{j=0}^{k} (-1)^j\binom{n}{k}\binom{k+1}{j+1}\frac{1}{g^{j+1}}
D_x^{n-k}\left(f\right)D_x^{k}\left(g^j\right)\tag{5}
\end{align*}
  and the claim follows.

Comment:


*

*In (4) we apply the formula (3)

*In (5) we exchange $k$ with $n-k$.

Let's look at a small example in order to see the  formula in action
Example: $D^2_x\left(\tan x\right)$
\begin{align*}
D_x^2\left(\frac{\sin x}{\cos x}\right)&=\sum_{k=0}^2\binom{2}{k}\left(D_x^{2-k}\sin    x\right)
\sum_{j=0}^k(-1)^j\binom{k+1}{j+1}\frac{1}{\cos^{j+1} x}\left(D_x^k\cos ^jx\right)\\
&=\binom{2}{0}\left(D_x^2\sin     x\right)\left((-1)^0\binom{1}{1}\frac{1}{\cos  x}D_x^0(1)\right)\\
&\qquad+\binom{2}{1}\left(D_x\sin x\right)\left((-1)\binom{2}{2}\frac{1}{\cos ^2x}\left(D_x \cos x\right)\right)\\
&\qquad+\binom{2}{2}\left(D_x^0 \sin x\right)\left((-1)\binom{3}{2}\frac{1}{\cos ^2x}\left(D_x^2  \cos x\right)\right.\\
&\qquad\qquad\qquad\qquad\qquad\qquad +\left.\binom{3}{3}\frac{1}{\cos^3x}\left(D_x^2\cos^2x\right)\right)\\
&=-\frac{\sin x}{\cos   x}+2\cos  x\left(\frac{\sin  x}{\cos  ^2x}\right)
+\sin x\left(3\frac{1}{\cos x}+\frac{2\sin^2 x - 2\cos ^2 x}{\cos ^3x}\right)\\
&=2\frac{\sin x}{\cos x}+2\frac{\sin ^3x}{\cos ^3x}\\
&=2\frac{\sin x}{\cos ^3x}\\
\end{align*}
  in accordance with Wolfram Alpha

