What is $\frac{d^n}{dx^n} \frac{e^{\lambda x}}{x}$?

Question

I was wondering whether there is an explicit way to say what the derivative of $\dfrac{d^n}{dx^n} \dfrac{e^{\lambda x}}{x}$ for $n \in \mathbb{N}_0$is, where we assume that $\lambda \neq 0$.

Answer 1

Since $\displaystyle(fg)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(k)}g^{(n-k)}$, setting $f(x)=\frac1x$ and $g(x)=e^{\lambda x}$ we obtain $$\frac{d^n}{dx^n}\left(\frac{e^{\lambda x}}{x}\right)=\sum_{k=0}^n {n \choose k} \frac{(-1)^k k!}{x^{k+1}} \cdot \lambda^{n-k}e^{\lambda x}=n!\lambda^n \frac{e^{\lambda x}}{x}\sum_{k=0}^n\frac{(-1)^k}{(n-k)!}(\lambda x)^{-k}.$$

Answer 2

Hint: Try a few derivatives and see a patern. Such a try motivates us to try an ansatz

$$\frac{d^n}{dx^n} \frac{e^{\lambda x}}{x} = \frac{P_n(x)}{x^{n+1}} e^{\lambda x}$$

where $P_n$ is a polynomial of degree $n$. If we take the derivative of this we can derive a recursion formula for $P_n$

$$P_{n+1}(x) = x P'_n(x) + (\lambda x - n - 1) P_n(x)$$

With $P_1(x) =\lambda x -1$ we find $P_2(x) = x^2\lambda^2 - 2 x\lambda +2$ and so on. It is easy to prove that $P_n$ is indeed a polynomial of degree $n$ with leading term $x^n\lambda^n$. It should also be possible to find a closed formula for $P_n$ by solving the recurence above.

Answer 3

Faa di Bruno formula: $$\frac{d^n}{dx^n}\frac{e^{\lambda x}}{x}=\sum_{i=0}^n{n \choose i} \frac{d^i}{dx^i}\frac{1}{x}\frac{d^{n-i}}{dx^{n-i}}e^{\lambda x}=\sum_{i=0}^n{n \choose i}(-1)^i\frac{i!}{x^{i+1}}\lambda^{n-i}e^{\lambda x}$$