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,
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n\cdot x^{(n-1)}$$
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n \cdot\dfrac {x^n}{x}$$

$$\frac{x^n}{x-1}=\frac{x^n-1}{x-1}+\frac1{x-1}=P(x)+\frac1{x-1}$$ 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 $\dfrac1{x-1}$,
$$\frac{(-1)(-2)\cdots(-n)}{(x-1)^{n+1}}.$$
