Tukukkk

I am trying to write a closed form for the $n$th derivative of $x log(x)$.

$frac{d}{dx} x log(x)= log(x)+1$

$frac{d^{2}}{dx^{2}}= frac{1}{x}$

and then from here there are no $log(x)$ terms ever again, it might as well be the $m$th derivative of $frac{1}{x}$.
So...do I need the ever-daunted step function? Or how can I describe this nth derivative in closed form?

$$ y'=ln x+1$$

And for $n>1$,

$$ y'' = 1/x =x^{-1}$$

$$y'''=-x^{-2}$$
$$y^{(4)} = 2 x^{-3}$$$$ y^{(n)} = (-1)^{n}(n-2)!x^{-n+1}$$

If $n$ is a positive integer, the function $h(n)=lfloor frac{1}{n} rfloor$ takes values $1$ for $n=1$ and $0$ for $n>1$. Therefore, using computations from @Mohammad Riazi-Kermani's answer, we have

$$
frac{d^n}{dx^n}(xln(x)) = h(n)(1+ln(x))+(1-h(n))(-1)^n(n-2)!x^{-n+1}
$$

for any $ngeqslant 1$.

IMHO, it is slightly annoying to have to deal with the special case $n=1$ but I don't think that the closed formula is more convenient. One may also argue that using the floor function is somehow cheating, since the floor function is just a (very convenient) way to disguise a piecewise function...

So basically, you are asking for a formula that is not piecewise-defined, right?

If so, one good way to do this is to use power series, since they are very easy to differentiate to arbitrary orders, even if the original function has a complicated mix of transcendental functions, as the terms are usually simply polynomial. To maximize the easiness, however, we might first want to make a substitution to "conjugate" the differentiation problem to a simpler one. Namely, we can use the Mercator series for $log(1 + x)$, which gives

$$log(1 + x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$$

We first rearrange this a bit, setting $x mapsto x - 1$, so that

$$log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} (x - 1)^n$$

thus

$$x log(x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x(x - 1)^n$$

and so the problem now reduces to differentiating the polynomial expression $x(x - 1)^n$, i.e. finding

$$frac{d^m}{dx^m} [x(x - 1)^n]$$

in terms of $m$. We can expand this using the binomial theorem to get

$$x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} x^{k+1}$$

and now we can differentiate $x^k$ $m$ times using the formula

$$frac{d^m}{dx^m} x^k = (k)_m x^{k-1}$$

where $(k)_m$ is the "falling factorial" given by $(a)_n = a(a-1)(a-2)cdots(a-n+1)$, which has $n$ factors. Note that if $m > k$, this is zero, and the derivatives thus go to zero.

Thus

$$frac{d^m}{dx^m} x(x - 1)^n = sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}$$

and the original is

$$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} left[sum_{k=0}^{n} binom{n}{k} (-1)^{n-k} (k+1)_m x^{k+1-m}right]$$

for which a rearrangement gives

$$frac{d^m}{dx^m} [x log(x)] = sum_{n=1}^{infty} sum_{k=0}^{n} binom{n}{k} frac{(-1)^{k+1} (k+1)_m}{n} x^{k+1-m}$$

This series will, of course, only converge in $x in (0, 2)$, I believe, but this is the closest thing to a "closed form" you will get even though it's not technically one, but moreover you can even vary $m$ smoothly, if you use the extension $(k+1)_m = frac{Gamma(k+2)}{Gamma(k-m-2)}$ and this will vary smoothly as well, so that you will have a continuous transition from $ln x$ to $frac{1}{x}$, and thus describes, or better, interpolates, the "transition" you ask about, and uses no ad-hoc step functions.

There is probably also some way to express the series using some kind of hypergeometric functions, but I don't feel like trying to rearrange it right now.