<s>[INST] 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.
 [/INST] Approach 1: When we calculate the first few derivatives, we have:
$$y'=-\frac{1}{x^2}\ln{x}+\frac{1}{x^2}=-\frac{1}{x^2}(\ln{x}-1).$$
$$y''=\frac{2}{x^3}\left(\ln{x}-1-\frac12 \right).$$
$$y^{(3)} = -\frac{3!}{x^4}\left(\ln{x}-1-\frac12 -\frac13\right).$$
so we can recognize the desired pattern.
Approach 2: We use the generalised Leibniz rule:
$$(fg)^{n} = \sum_{k= 0}^n \binom{n}{k} f^{(n-k)}(x) g^{(k)}(x)$$
with $f(x) = 1/x, g(x) = \log x$
to obtain:
$$\left(\frac{\log x}{x}\right)^{(n)}= \sum_{k= 0}^n \binom{n}{k} \log^{(k)}(x) \left(\frac{1}{x}\right)^{(n-k)}(x)$$
$$= \sum_{k= 0}^n \binom{n}{k} \log^{(k)}(x) \frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$=  \log(x) \frac{(-1)^{n}n!}{x^{n+1}} + \sum_{k= 1}^n \binom{n}{k} \left(\frac{1}{x}\right)^{(k-1)} \frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$= \log(x) \frac{(-1)^{n}n!}{x^{n+1}} +  \sum_{k= 1}^n \binom{n}{k} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$= \log(x) \frac{(-1)^{n}n!}{x^{n+1}} +  \sum_{k= 1}^n \frac{n!}{k!(n-k)!} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} +   \sum_{k= 1}^n \frac{n!}{k(k-1)!(n-k)!} \frac{(-1)^{k-1}(k-1)!}{x^{k}}\frac{(-1)^{n-k}(n-k)!}{x^{n-k+1}}$$
$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} +   \sum_{k= 1}^n \frac{n!}{k} \frac{(-1)^{k-1}}{x^{k}}\frac{(-1)^{n-k}}{x^{n-k+1}}$$
$$=\log(x) \frac{(-1)^{n}n!}{x^{n+1}} +   \sum_{k= 1}^n \frac{n!}{k} \frac{(-1)^{n-1}}{x^{n+1}}$$
$$ = \log(x) \frac{(-1)^{n}n!}{x^{n+1}} - \frac{n!(-1)^n}{x^{n+1}} \sum_{k= 1}^n \frac{1}{k}$$
$$= \frac{n!(-1)^n}{x^{n+1}} \left(\log(x) - 1 - \frac{1}{2} - \dots - \frac{1}{n}\right)$$
Approach 3
Induction: For $n=0$, the formula holds.
Assume it holds for $n-1$, and then show it holds for $n$, by taking the derivative and rearranging terms.
Other/similar approaches: n-th derivative of $\frac{\ln x}{x}$.
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