I'm doing it completely wrong, I'm sure, but I'll still show my attempt:
$n^\text{th}$ order derivative of $x^n$ is $n!$ and of $\ln x$ is $(-1)^{(n-1)} (n-1)! x^{-n}$
So, using Leibnitz rule I got $n^\text{th}$ order derivative of $x^n \ln x$ as
$$n!\ln x + n(n-1) \cdots 1 \cdot x \frac1x + \cdots + x^n \frac{1}{x^n} (-1)^{(n-1)} $$
But the form I'm supposed to get is $n! \left[ \ln x +1 + \frac12 + \cdots + \frac1n\right]$
Please help!
In this case the Leibniz rule leads to a complicated sum over many factorials and a minus sign.
To see why the $n^th$ order differentiation of $x^n log(x)$ leads to the required sum, let us avoid the Leibniz rule and perform the differentiation one step at a time, using the product rule.
order 1: $\quad$ $nx^{n-1}log(x) + x^{n-1}$
order 2: $\quad$ $n(n-1)x^{n-2}log(x) + nx^{n-2} + (n-1)x^{n-2}$
order 3: $\quad$ $n(n-1)(n-2)x^{n-3}log(x) + n(n-1)x^{n-3} + n(n-2)x^{n-3} + (n-1)(n-2)x^{n-3}$
At this stage the result still looks somewhat messy, but the pattern becomes clear if we reorganize the above results as follows:
order 1: $\quad$ $nx^{n-1} [log(x) + 1/n]$
order 2: $\quad$ $n(n-1)x^{n-2}[log(x) + 1/(n-1) + 1/n]$
order 3: $\quad$ $n(n-1)(n-2)x^{n-3}[log(x) + 1/(n-2) + 1/(n-1) + 1/n]$
Hence we are now in the position to extend the result to the $n^{th}$ order derivative. We obtain:
order n: $\quad$ $n! [log(x) + 1/1 + 1/2 + 1/3 ...... + 1/n]$
This is indeed the required result.
