<s>[INST] Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$. Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$.
My Attempt:
$$y=\dfrac {x^n}{x-1}$$
$$y=x^n\cdot(x-1)^{-1}$$
Differentiating both sides,
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n\cdot x^{(n-1)}$$
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n \cdot\dfrac {x^n}{x}$$
 [/INST] $f(x) = \frac{x^n}{x-1} = \frac{x^n-1+1}{x-1} = x^{n-1} + x^{n-2} + \ldots + x + 1 + \frac{1}{x-1} \Rightarrow$
$f^{(n)}(x) = \left(x^{n-1} + x^{n-1} + \ldots + x + 1\right)^{(n)} + \left(\frac{1}{x-1}\right)^{(n)} = 0 + (-1)^n\frac{n!}{(x-1)^{n+1}} = (-1)^n\frac{n!}{(x-1)^{n+1}}$.
</s>