Show that $\frac{d^n}{dx^n}(\frac{\log x}{x})=(-1)^n\frac{n!}{x^{n+1}}(\log x-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n})$ Show that $\frac{d^n}{dx^n}\left(\frac{\log x}{x}\right)=(-1)^n\frac{n!}{x^{n+1}}\left(\log x-1-\frac{1}{2}-\frac{1}{3}-\ldots -\frac{1}{n}\right)$
I am not allowed to use induction. I do not know how to approach this. Any help would be appreciated.

Another proof
Let $f(x) = \frac{\log(x)}{x}$, then for $x, y > 0$
$$f(x y) = \frac{\log(x)+ \log(y)}{x y} = \frac{f(x)}{y} + \frac{f(y)}{x}$$
Derivating $n$ times with respect to $y$ yields
$$x^n f^{(n)}(x y) = f(x) \left(\frac{1}{y}\right)^{(n)} + \frac{f^{(n)}(y)}{x}$$
Now replacing $y$ by $1$ yields
$$x^n f^{(n)}(x) = f(x) (-1)^n n! + \frac{f^{(n)}(1)}{x}$$
hence
$$f^{(n)}(x) = (-1)^n n! \frac{\log(x)}{x^{n+1}} + \frac{f^{(n)}(1)}{x^{n+1}}$$
The value of $f^{(n)}(1)$ is missing, it can be obtained by the Taylor series of
$$f(1+h) = \frac{\log(1+h)}{1+h}$$
used by Professor Vector, but there is another way: rewrite the above formula as
$$f^{(n)}(x) = (-1)^{n}n! \frac{\log(x) - s_n}{x^{n+1}}$$
We have $s_0=0$ and we claim that $s_{n+1} - s_n = \frac{1}{n+1}$ from which it follows that
$$s_n = s_n - s_0 = \sum_{k=0}^{n-1}(s_{k+1}-s_k) = \sum_{k=0}^{n-1}\frac{1}{k+1} = 1 + \frac{1}{2} +\cdots+\frac{1}{n}$$
To prove our claim, we compute
$$(-1)^{n+1}(n+1)! \frac{\log(x) - s_{n+1}}{x^{n+2}} = f^{n+1}(x)
= (f^{(n)}(x))^\prime
=(-1)^{n}n! \frac{1 -(n+1)(\log(x) - s_n)}{x^{n+2}}$$
It follows that $s_{n+1} = \frac{1}{n+1} + s_n$, QED
