# N:th derivative of exp(x)/x type function?

1. Sep 26, 2011

### winterfors

I have a function
$$f(x) = \exp (x)\prod\limits_{j = 1}^n {{{(x - {y_j})}^{ - {z_j}}}}$$
of which I need to find an expression of the k:th derivative with respct to x:
$$\frac{{{\partial ^k}}}{{\partial {x^k}}}f(x)$$ .

I have been able to make a conjecture that seems to be correct
$$\frac{{{\partial ^k}}}{{\partial {x^k}}}f(x) = f(x)k!\sum\limits_{{\bf{l}} \in {\Lambda _k}} {\frac{1}{{{l_1}!}}{{\left( {1 + \sum\limits_{j = 1}^n {\frac{{{z_j}}}{{{y_i} - x}}} } \right)}^{{l_1}}}\prod\limits_{a = 2}^k {\frac{1}{{{l_a}!}}{{\left( {\frac{1}{a}\sum\limits_{j = 1}^n {\frac{{{z_j}}}{{{{({y_i} - x)}^a}}}} } \right)}^{{l_a}}}} }$$

where ${\Lambda _k}$ is the set of k-length vectors of positive integers so that ${\bf{l}} \cdot {[1,2,...,k]^{\rm{T}}} = k$ $\forall {\bf{l}} \in {\Lambda _k}$.

I have no idea of how to prove it, but I'm sure someome must have looked at such a simple case before. Does anyone have an idea of how to prove it, or know of a reference for this kind of formula?

2. Sep 26, 2011

### Mute

The Liebniz rule deals with derivatives of products. Writing your function as $f(x) = \exp(x)g(x)$, the rule gives

$$f^{(k)}(x) = \exp(x) \sum_{\ell=0}^k \frac{k!}{\ell!(k-\ell)!}g^{(\ell)}(x),$$

so you need to figure out the derivative of the g(x) function, which you can try to do using the Liebniz rule again. You can probably help churn out your formula this way.

3. Sep 27, 2011

### winterfors

Well, using the general Liebniz rule you can quite easily get to
$$\frac{{{\partial ^k}}}{{\partial {x^k}}}f(x) = f(x)k!\sum\limits_{{\bf{m}} \in \Theta _k^{n + 1}} {\frac{1}{{\prod\limits_{j = 1}^{n + 1} {{m_j}!} }}} \prod\limits_{j = 1}^n {\frac{{({z_j} - 1 + {m_j})!}}{{({z_j} - 1)!}}} {({y_j} - x)^{ - {m_j}}}$$

where $\Theta _k^{n + 1}$ is the set of all n+1 length vectors of positive integers such that ${m_1} + {m_2} + \cdots + {m_{n + 1}} = k$.

How to get from this sum to the conjecture above seems a bit more tricky...