Is there a formula for the nth derivative of $\frac{1}{f(x) +a}$? Consider $\dfrac{1}{f(x) +a}$, where $a$ is a real number $f(x)$ is a smooth real function. I wonder is there an explicit formula for the $n^{\text{th}}$ derivative of $\dfrac{1}{f(x)+a}$.
 A: I played with this a bit; It seems that the $n$th derivative satisfies an identity which is somehow similar to the binomial theorem. Let
$$
g(x)=\frac{1}{f(x)+a}\tag{1}
$$
Suppose that $f(x)\not=-a$. Multiply both sides of $(1)$ by $f(x)+a$ to get
$$
g(x)f(x)=1-ag(x)\tag{2}
$$
Taking the first derivative of $(2)$ with respect to $x$ gives
$$
g'(x)f(x)+g(x)f'(x)=-ag'(x)\tag{*}
$$
Taking the second derivative of $(2)$ with respect to $x$ gives
$$
g''(x)f(x)+2g'(x)f'(x)+g(x)f''(x)=-ag''(x)
$$
Taking the third derivative of $(2)$ with respect to $x$ gives
$$
g'''(x)f(x)+3g''(x)f(x)+3g'(x)f''(x)+g(x)f'''(x)=-ag'''(x)
$$
Taking the fourth derivative of $(2)$ with respect to $x$ gives
$$
g^{(4)}(x)f(x)+4g'''(x)f'(x)+6g''(x)f''(x)+4g'(x)f'''(x)+g(x)f^{(4)}(x)=-
ag^{(4)}(x)
$$
and so on. In general, if $n\in\mathbb{N}$ where $\mathbb{N}$ is the set of natural numbers including $0$ and with the convention that $f^{(0)}(x)=f(x)$, then the $n$-th derivative of $(2)$ satisfies the following identity
$$
\sum_{k=0}^n\binom{n}{k}g^{(n-k)}(x)f^{(k)}(x)=-ag^{(n)}(x)\tag{3}
$$
where the left hand side of identity $(3)$ is the $n$th derivative of $f(x)g(x)$.
Now from $(*)$, we have
$$
g'(x)=-f'(x)(g(x))^2
$$
We already discovered that
$$
(g(x)f(x))^{(n)}=\sum_{k=0}^n\binom{n}{k}g^{(n-k)}(x)f^{(k)}(x)
$$
So now we have
$$
\begin{align}
g^{(n+1)}(x)&=\frac{\mathrm{d}^n}{{\mathrm{d}x}^n}g'(x)\\
&=\frac{\mathrm{d}^n}{{\mathrm{d}x}^n}\left(-f'(x)(g(x))^2\right)\\
&=\sum_{k=0}^n\binom{n}{k}(-f'(x))^{(n-k)}\left((g(x))^2\right)^{(k)}\\
&=-\sum_{k=0}^n\left(\binom{n}{k}f^{(n-k+1)}(x)
\sum_{j=0}^k\binom{k}{j}g^{(k-j)}(x)g^{(j)}(x)
\right)
\end{align}
$$
A: As I mentioned in my comment, you may apply the Faà di Bruno's formula, which expands the $n$-th derivative of a composite function in combinatoric way.
In order to describe the formula, it is convenient to prepare some notations. Let $\lambda = (\lambda_1, \cdots, \lambda_n)$ be an $n$-tuple of non-negative integers. Then


*

*Write $\lambda \vdash n$ if it satisfies $\sum_{i=1}^{n} i \lambda_i = n$.

*Write $|\lambda| = \sum_{i=1}^{n} \lambda_i$.


Then the Faà di Bruno's formula tells that
$$ (g \circ f)^{(n)} = \sum_{\lambda \vdash n} \frac{n!}{\lambda_1! (1!)^{\lambda_1} \cdots \lambda_n! (n!)^{\lambda_n}} (g^{(|\lambda|)} \circ f) \prod_{i=1}^{n} ( f^{(i)} )^{\lambda_i} \tag{*} $$
where the sum is taken over all $n$-tuples $\lambda$ satisfying $\lambda \vdash n$. Now plugging $g(x) = \frac{1}{a+x}$, we know a simple formula for $g^{(k)}$ and hence we obtain
$$\frac{d^n}{dx^n} \left( \frac{1}{a+f(x)} \right)
= \sum_{\lambda \vdash n} \frac{n!}{\lambda_1! (1!)^{\lambda_1} \cdots \lambda_n! (n!)^{\lambda_n}} \left( \frac{(-1)^{|\lambda|} |\lambda|!}{(a + f(x))^{|\lambda|+1}} \right) \prod_{i=1}^{n} \big( f^{(i)}(x) \big)^{\lambda_i}
$$
This egregious expression has no hope of being simplified further unless $f$ demonstrates a very nice algebraic property under differentiation.
