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 $$u=1-x \Longrightarrow (u)^{-2}\cdot( -1)= \frac{-1}{(1-x)^2}$$

$$\frac d{dx} (1-x)^{-1} = -1\cdot (1 - x)^{-2} \cdot \underbrace{\frac{d}{dx}(1-x)}_{\large =\,-1}$$ 
$$ = -1\cdot -1\cdot (1 - x)^{-2}= \frac{1}{(1-x)^2}$$
