Find $f^{(n)}(1)$ where $f(x)={1\over x(2-x)}$. Find $f^{(n)}(1)$ where $f(x)={1\over  x(2-x)}$. 
What I did so far: 
$f(x)=(x(2-x))^{-1}$. $f'(x)=-(x(2-x))^{-2}[2-2x]$ $f''(x)=2(x(2-x))^{-3}[2-2x]^2+2(x(2-x))^{-2}$. It confuses me a lot. I know I have to determine where I have an expression multiplied by $2-2x$ and where not, and what is its form. I would appreciate your help.  

Since for any $a\neq 0$:
$$\frac{1}{x(a-x)} = \frac{1}{a}\left(\frac{1}{x}+\frac{1}{a-x}\right)\tag{1}$$
we have:
$$ \frac{d^n}{dx^n}\frac{1}{x(a-x)} = \frac{n!}{a}\left(\frac{(-1)^n}{x^{n+1}}+\frac{1}{(a-x)^{n+1}}\right).\tag{2}$$
