# 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.

-
Look more closely at what Alpha gives you back - it's not quite the first half of what you've written. What you've written is actually correct, but try expanding out the result and you may see where Alpha is coming from. – Steven Stadnicki Jan 9 '13 at 16:29
I have several times with no luck. – user56699 Jan 9 '13 at 16:30
$$-(1-x^{-1})^{-2}x^{-2}=\frac{-1}{x^2(1-x^{-1})^2}=\frac{-1}{x^2\left(1-\frac1x‌​\right)^2}=\frac{-1}{\left(x\left(1-\frac1x\right)\right)^2}=\frac{-1}{(x-1)^2}$$ – Brian M. Scott Jan 9 '13 at 16:33
@BrianM.Scott Why did the x^2 turn into just an x in the third step? – user56699 Jan 9 '13 at 16:35
I don't know how you knew to use those specific and very odd manipulations but I did something more intuitive and it just doesn't work. I took $(x^{-1} - 1$ and squared it getting $x^{-2} - 2x^{-1} + 1$ which is very bad and wrong. I then multiply that by $x^2$ and get $(1-2x+x^2)^{-1}$ – user56699 Jan 9 '13 at 16:38

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 $(3)$

-

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: $$\log \frac{f(x)}{h(x)}=\log f(x)-\log h(x)\\ \textbf{ If } g(x)=\log f(x) \textbf{ then } \\ g'(x)=\frac{f'(x)}{f(x)} \Leftrightarrow \\ f'(x)=f(x)g'(x)$$

-
I have no idea why you introduced logs into this, what L means, what L! is or really anything that you did. It might be neater but I don't think it is anything we were ever taught. – user56699 Jan 9 '13 at 16:39
Pls see edits above – Alex Jan 9 '13 at 16:44
I really can't decipher what you wrote at all. I think you wrote teh chain rule out maybe and you are saying that you took the logs of both sides but I am not sure when or how they went away. – user56699 Jan 9 '13 at 16:49
I didn't use Chain rule here because it is not the only way of solving this problem. Yes, I took logs of both sides and then differentiated the function:$(\log x)'_{x}=\frac{1}{x}$ – Alex Jan 9 '13 at 16:51
I am not familiar with that at all don't you have to raise everything to e or something like that to get rid of log? – user56699 Jan 9 '13 at 17:08

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.

-
+1 Nice...I remember this problem ;-) – amWhy Feb 20 '13 at 0:28

No, you're right. You just expressed it a little diffently. Remember that $a^xb^x=(ab)^x$

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

-