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.
 A: 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

