<s>[INST] 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?
 [/INST] $\frac {x}{1-x} = \frac {x}{1-x}+1-1 = \frac {1}{1-x} - 1\\
\frac {d}{dx} (1-x)^{-1}-1= (1-x)^{-2}\\
\frac {d^n}{dx^n} (1-x)^{-1}-1= n!(1-x)^{-(n+1)}\\
$
With what you have above, you do change the sign of your expression in the first step.  I don't know if that is intentional.
at $\frac {d}{dx} x\frac {1}{x-1} = \frac {1}{x-1} - \frac {x}{(x-1)^2}$ combine the terms.
$\frac {1}{x-1} - \frac {x}{(x-1)^2} = \frac {x-1 - x}{(x-1)^2} = \frac {-1}{(x-1)^2}$
</s>