# 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.

• "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 $k>0$." But this is not true, for example $h''(0)=-2$.
– Did
Feb 24, 2016 at 14:07
• I calculate $h^{\prime\prime}(x)=-\frac{2x}{(1+x^2)^2}$. This is zero. Feb 24, 2016 at 14:44
• Yeah, and your computation of $h''$ is wrong (you wrote $h'$, not $h''$).
– Did
Feb 24, 2016 at 14:47
• Yes, $h^{(k)}(0)=0$ if and only if $k$ is odd. CAS are not necessary to see this, since $h$ is an even function.
– Did
Feb 24, 2016 at 15:15
• And from $h(x)=\sum_{k=0}^\infty(-x^2)^k=\sum_{k=0}^\infty(-1)^kx^{2k}$ you get all the other derivatives by comparing it with the Maclaurin series. Feb 24, 2016 at 16:48

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

• Just found your answer. Of course I didnt try to follow your steps yet. Seems to me to be what I wasnt able to calculate until a successful end. Am quite excited to see whether I can apply it in my work. Mar 31, 2016 at 13:46
• @WolfgangTintemann: In fact, here are three useful formulas stated: Hoppes formula $D_x^n(f(g))$, $D_x^n\left(\frac{f}{g}\right)$ and $D_x^n\left(\frac{1}{g}\right)$. I've added an answer to this question where you can also see Hoppe's formula in action. Maybe it is helpful for you. Mar 31, 2016 at 14:07
• What I am thinking about now is to interchange the sums over $k$ and over $j$ in formula (5) to get a sum over the product of the binomial coefficients where the common $k$-entry is summed. Wouldnt surprise me if one could then find a simplification of this sum in Prof Goulds book. At least that would lead to a double-sum-less compact formula. Mar 31, 2016 at 15:11
• @WolfgangTintemann: Nice idea, but I think it's not feasible. We can hope to get rid of a sum, if we could put one of the binomial coefficients out of the inner sum. But this is not the case as far as I see. Another obstacle is that $k$ and $j$ are closely interwoven in the expression $D_x^k(g^j)$. Regards, Mar 31, 2016 at 15:26