Is there a formula for the nth derivative of $\frac{1}{f(x) +a}$?

Question

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 $\dfrac{1}{f(x)+a}$.

Answer 1

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 $$ \begin{align} g^{(n+1)}(x)&=\frac{\mathrm{d}^n}{{\mathrm{d}x}^n}g'(x)\\ &=\frac{\mathrm{d}^n}{{\mathrm{d}x}^n}\left(-f'(x)(g(x))^2\right)\\ &=\sum_{k=0}^n\binom{n}{k}(-f'(x))^{(n-k)}\left((g(x))^2\right)^{(k)}\\ &=-\sum_{k=0}^n\left(\binom{n}{k}f^{(n-k+1)}(x) \sum_{j=0}^k\binom{k}{j}g^{(k-j)}(x)g^{(j)}(x) \right) \end{align} $$

Answer 2

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.