I have a function(adsbygoogle = window.adsbygoogle || []).push({});

[tex]f(x) = \exp (x)\prod\limits_{j = 1}^n {{{(x - {y_j})}^{ - {z_j}}}} [/tex]

of which I need to find an expression of thek:th derivative with respct tox:

[tex]\frac{{{\partial ^k}}}{{\partial {x^k}}}f(x)[/tex] .

I have been able to make a conjecture that seems to be correct

[tex]\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}}}} } [/tex]

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

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?

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

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