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,
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n\cdot x^{(n-1)}$$
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n \cdot\dfrac {x^n}{x}$$
 A: Denote $t = x-1$, then 
$$ y = \frac{(t+1)^n}{t} = p(t) + \frac{1}{t}$$
where $p$ is a polynomial of degree $n-1$ (in x), so $p^{(n)}(t) = 0$. Therefore
$$\frac{d^ny}{dx^n} = \frac{d^n}{dx^n}\left(\frac{1}{x-1}\right).$$
It should be easy from here.
A: $f(x) = \frac{x^n}{x-1} = \frac{x^n-1+1}{x-1} = x^{n-1} + x^{n-2} + \ldots + x + 1 + \frac{1}{x-1} \Rightarrow$
$f^{(n)}(x) = \left(x^{n-1} + x^{n-1} + \ldots + x + 1\right)^{(n)} + \left(\frac{1}{x-1}\right)^{(n)} = 0 + (-1)^n\frac{n!}{(x-1)^{n+1}} = (-1)^n\frac{n!}{(x-1)^{n+1}}$.
A: $$\frac{x^n}{x-1}=\frac{x^n-1}{x-1}+\frac1{x-1}=P(x)+\frac1{x-1}$$ where $P$ is a polynomial of degree $n-1$. The $n^{th}$ derivative of this polynomial vanishes and we are left with the $n^{th}$ derivative of $\dfrac1{x-1}$,
$$\frac{(-1)(-2)\cdots(-n)}{(x-1)^{n+1}}.$$
A: Recall the general Leibniz rule, $(fg)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(n-k)} g^{(k)}$, which can be proven by induction. $g^{(k)}$ stands for the $k$'th derivative of $g$. Set $f=x^n$, so $f^{(n-k)}=n\cdot(n-1)\cdot\ldots[n-(n-k-1)]\;x^{n-(n-k)}=\frac{n!}{k!}x^{k}$. Then, with $g=(x-1)^{-1}$, $g^{(k)}=(-1)\cdot(-2)\cdot\ldots(-k)\;(x-1)^{-(k+1)}=\frac{(-1)^kk!}{(x-1)^{k+1}}$. So, finally, $y^{(n)}=\sum_{k=0}^n {n \choose k} \frac{n!}{k!}x^{k} \frac{(-1)^kk!}{(x-1)^{k+1}}=\frac{n!}{x-1}\sum_{k=0}^{n}\binom{n}{k}\left(\frac{x}{1-x}\right)^{k}=-\frac{n!}{\left(1-x\right)^{n+1}}$.
