Nth Derivative of $\frac{x}{x-1}$ So starting off, rewrite $x*\dfrac1{x-1}$. $d/dx=\dfrac{d}{dx}(x)*\dfrac{1}{x-1}+\dfrac{d}{dx}\dfrac{1}{x-1}*x$.
$\dfrac{d}{dx}(x)=1$. $x^{n}=nx^{n-1}$.
$\dfrac{1}{x-1}=(x-1)^{-1}$.
$\dfrac{d}{dx}(\dfrac{1}{x-1})=(-1)(x-1)^{-2}=\dfrac{-1}{(x-1)^{2}}$.
Combining them both we get $\dfrac{1}{x-1}+\dfrac{-1}{(x-1)^{2}}*x$.
$\dfrac{d}{dx}=\dfrac{1}{x-1}+\dfrac{-x}{(x-1)^{2}}$.
Now to find $\dfrac{d^2y}{d^2x}$, using product rule and sum rule
$\dfrac{d}{dx}\dfrac{1}{x-1}+\dfrac{d}{dx}-x\dfrac{1}{(x-1)^2}+\dfrac{d}{dx}\dfrac{1}{(x-1)^{2}}x$, so anyway doing it we get
$\dfrac{-1}{(x-1)^2}+\dfrac{-1}{(x-1)^2}+\dfrac{-2x}{(x-1)^{3}}$.
Simplifying $\dfrac{d^2y}{d^2x}=\dfrac{-2}{(x-1)^2}+\dfrac{-2x{(x-1)^{3}}$; and finally $\dfrac{d^3y}{d^3x}$. $\dfrac{d}{dx}\dfrac{-2}{(x-1)^2}+\dfrac{d}{dx}-2x\dfrac{1}{(x-1)^{3}}+\dfrac{d}{dx}\dfrac{1} {(x-1)^{3}}*(-2x)$,which gives $\dfrac{4}{(x-1)^3}+\dfrac{-2}{(x-1)^3}+\dfrac{6x{(x-1)^4}$; simplyfing we get
$\dfrac{2}{(x-1)^3}+\dfrac{6x}{(x-1)^4}$; now there is supposed to be a pattern here but i don't see it.
is there something wrong with my answer?

How about
$$f(x)=\frac x{x-1}=\frac1{x-1}+1$$
Therefore:
$$\frac d{dx}f(x)=-1\frac1{(x-1)^2}\\\frac{d^2}{dx^2}f(x)=-1\cdot-2\frac1{(x-1)^3}\\\frac {d^3}{dx^3}f(x)=-1\cdot -2\cdot -3\frac1{(x-1)^4}\\\vdots\\\frac{d^n}{dx^n}f(x)=\frac{(-1)^n n!}{(x-1)^{n+1}}$$
One may use factorial power $u^{(v)}$ to get:
$$\boxed{\frac{d^n}{dx^n}\frac x{x-1}=\frac{(-1)^{(n)}}{(x-1)^{n+1}},n\ge 1}$$
