<s>[INST] Finding the derivative to nth order How to find $$\frac{d^ny}{dx^n}$$ 
of $$y=\frac{x}{lnx-1}$$
Appreciated advance
 [/INST] By applying the quotient rule we have $$\frac{dy}{dx} = \frac{1 \times (\ln x -1)- \frac{1}{x}x}{(\ln x - 1)^2} = \color{blue}{\frac{\ln x -2}{(\ln x - 1)^2}}$$
Now you can apply the quotient rule again to get $$\frac{d^2y}{dx^2} = \frac{\frac{1}{x} \times (\ln x -1)^2 - 2(\ln x -1) \frac{1}{x} \times (\ln x - 2)}{(\ln x - 1)^4} =\frac{\frac{1}{x}(\ln x -1)((\ln x-1) -2(\ln x -2))}{(\ln x - 1)^4} = \frac{\frac{1}{x}((\ln x-1) -2(\ln x -2))}{(\ln x - 1)^3}=\frac{\frac{1}{x}(\ln x -2\ln x+3)}{(\ln x - 1)^3}=\frac{(-\ln x+3)}{x(\ln x - 1)^3} = \color{blue}{\frac{-(\ln x-3)}{x(\ln x - 1)^3}}$$
Now do you see a pattern developing it is $$\color{blue}{\frac{d^ny}{dx^n}=\frac{(-1)^{n+1}(\ln x -(n+1))}{x^{n-1}(\ln x - 1)^{n+1}}}$$
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