Simple chain rule application $y = (1-x^{-1})^{-1}$ I am not sure what is going wrong here. I have been doing other applications of the chain rule to cross check that I understand it properly ,but I still do not get a correct answer on this problem while I do on all others.
$$y = (1-x^{-1})^{-1}$$
$$y' = -(1-x^{-1})^{-2} \cdot x^{-2}$$
This is wrong and it  gives a wrong answer, according to wolfram and my book the answer should just be the first part which breaks the chain rule and I do understand why this is acceptable in this specific case but no others.
 A: You applied the chain rule correctly, and obtained "a" correct answer; the discrepancy you find is "simply" due to the fact that your answer  can be simplified. (No pun intended!)
Expanding your correct result: $\;\;y' = -(1-x^{-1})^{-2}x^{-2};\;$ we simplify as follows: $$=\frac{-1}{x^2(1-x^{-1})^2}\tag{1}$$
$$=\frac{-1}{x^2\left(1-\frac1x‌​\right)^2}\tag{2}$$ 
$$=\frac{-1}{\left(x\left(1-\frac1x\right)\right)^2}\tag{3}$$ $$=\frac{-1}{(x-1)^2}\tag{4}$$  $$= -(x-1)^{-2}\tag{5}$$
So your computation/answer is equivalent to $(5)$.

Note: $(2) \to (3)\;\;$ uses the fact that $\;x^ny^n = (xy)^n,\;\text{so}\; (2)$ is equivalent to $(3)$
A: There is a neater way of handling this. Denote $\log f(x)=Lf(x)$.
$$
f(x)=\bigg(1-\frac{1}{x}\bigg)^{-1}\\
Lf(x)=- \log \bigg(1-\frac{1}{x}\bigg)=\log x -\log (x-1)\\
L'f(x)=\frac{1}{x}-\frac{1}{x-1}=-\frac{1}{x(x-1)}\\
f'(x)=-\frac{f(x)}{x(x-1)}=-\frac{1}{(x-1)^2}
$$
EDIT: here I used the following properties of logarithm function:
$$
\log \frac{f(x)}{h(x)}=\log f(x)-\log h(x)\\
 \textbf{ If } g(x)=\log f(x) \textbf{ then } \\
g'(x)=\frac{f'(x)}{f(x)} \Leftrightarrow \\
f'(x)=f(x)g'(x)
$$
A: No, you're right. You just expressed it a little diffently. Remember that $a^xb^x=(ab)^x$
$-(1-x^{-1})^{-2}x^{-2}=-((1-x^{-1})x)^{-2}=-(x-1)^{-2}=-\frac{1}{(x-1)^2}$
A: Set $f(x)=u^{-1}$ wherein $u=1-\frac{1}{x}$. So according to chain rule we have: $$\frac{df(x)}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$$ Here, $$\frac{df}{du}=-1\times u^{-1-1}=-u^{-2}$$ and $$\frac{du}{dx}=-(x^{-1})'=x^{-2}$$ So you have your answer.
