General expression for the derivative?

[rsq_a]

I'm trying to find the general form for the nth derivative of
$$f(x) = \frac{1}{x^m \log x}$$

where m can be anything (set m = 1 for instance). For ease, you can take m to be integral.

It sounds surprisingly simple, but the most I've been able to say is
$$f^{(n)}(x) = (-1)^n x^{-(m+n)} \sum_{k=0}^n a_{k, n} [\log(x)]^{-k}$$

where the coefficients satisfy
$$a_{k,n} = [m + (n-1)] a_{k, n-1} + (k-1) a_{k-1, n-1}$$

for 0 < k < n, and with $$a_{0, n} = (m + n - 1)!/(m-1)!$$ and $$a_{n,n} = n!$$

Unfortunately, I was hoping to get a general form for the coefficients. Does anyone know a trick?

 

[fresh_42]

There is no trick. The Leibniz rule is responsible for the coefficients and you can't change that.