Derivative of $\frac1{1-x}$ Why is this not correct:
$$
\frac{1}{1-x}= (1-x)^{-1}
$$
now use chain-rule which gives: $(1-x)^{-2}$ times derivative of $(1-x)$ which is $-1$
so
$$
-1\cdot (1-x)^{-2}= \frac{-1}{(1-x)^2}
$$
why is this incorrect? 
Because if I use quotient rule on $1/(1-x)$ I get
$$
\frac{0 \cdot (1-x) - 1\cdot -1}{(1-x)^2}= \frac{1}{(1-x)^2}.
$$
So why do I get with using chain rule on $(1-x)^{-1}$ a different answer?
$$
\frac{d}{dx} (1-x)^{-1}= \frac{d}{du} (u)^{-1} \cdot \frac{d}{dx} (u),$$
 with $$u=1-x \Longrightarrow (u)^{-2}\cdot( -1)= \frac{-1}{(1-x)^2}$$

Remember $\frac{d}{dx}x^n=nx^{n-1}$, so in the first case you should have 
$$\frac{d}{dx}(1-x)^{-1}=(-1)(1-x)^{-2}\left(\frac{d}{dx}(1-x)\right)=(-1)(1-x)^{-2}(-1)=\frac{1}{(1-x)^2}$$
