$n^\text{th}$ derivative?

Let $y=\dfrac{\ln x}{x}$ . From the first few derivatives of $y$ we can deduce that: $\displaystyle \frac{d^ny}{dx^n} = \frac{(-1)^n n!}{x^{n+1}} \left(\ln x - \sum_{k=1}^n \frac{1}{k} \right)$ . Let us prove that it is true for all $n \ge 1$ by induction.

1. For n = 1, $\begin{aligned} \frac{dy}{dx} & = \frac{1}{x^2} - \frac{\ln}{x^2} = \frac{(-1)^11!}{x^{1+1}}\left(\ln x - \frac{1}{1} \right) \end{aligned}$ . Therefore, it is true for $n=1$ .

2. Assume it is true for $n$ , then:

$\begin{aligned} \quad \quad \frac{d^{n+1}y}{dx^{n+1}} & = \frac{d}{dx} \left( \frac{d^{n}y}{dx^{n}} \right) \\ & = \frac{d}{dx} \left(\frac{(-1)^n n!}{x^{n+1}} \left(\ln x - \sum_{k=1}^n \frac{1}{k} \right) \right) \\ & = \frac{(-1)^n n!(-1)(n+1)}{x^{n+2}} \left(\ln x - \sum_{k=1}^n \frac{1}{k} \right) + \frac{(-1)^n n!}{x^{n+2}} \\ & = \frac{(-1)^{n+1}(n+1)!}{x^{n+2}} \left(\ln x - \sum_{k=1}^n \frac{1}{k} - \frac{1}{n+1} \right) \\ & = \frac{(-1)^{n+1}(n+1)!}{x^{(n+1)+1}} \left(\ln x - \sum_{k=1}^{n+1} \frac{1}{k} \right) \end{aligned}$

$\quad$ So, it is also true for $n+1$ , therefore it is true for all $n \ge 1$ .