# Derivative of $y= (1-x^{-1})^{-1}$

I am trying to find the derivative of $y= (1-x^{-1})^{-1}$ to make it simple I use the chain rule and

$$u = 1-x^{-1}$$

$$y = u^{-1}$$

$$y' = -1u^{-2}u' = -1(1-x^{-1})^{-2}(x^{-2})$$

The problem is that this is not what my book gets and I am wondering if there is some strange power rule I overlooked or some other way to do this problem that gives me a different answer, or maybe I just need to factor is out somehow.

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I do not see the difference. –  user138246 May 5 '12 at 18:01
I'm sorry, I misread the question. –  David Mitra May 5 '12 at 18:04
Jordan, when you subsituted 'back' for $u^{-2}$, you forgot the $\;^{-2}$ part. –  copper.hat May 5 '12 at 18:10
I have that on my paper, I just typed it up wrong. –  user138246 May 5 '12 at 18:10
Most likely, another case where the book gives a simplified answer: $$-1(1+x^{-1})^{-2}x^{-2}= {-1\over (1-{1\over x})^2}\cdot {1\over x^2} = {-1\over x^2\cdot(1-{1\over x})^2} = {-1\over \bigl( x(1-{1\over x})\bigr)^2} = {-1\over (x-1 )^2} .$$ –  David Mitra May 5 '12 at 18:17
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$$y = \left( 1 - x^{-1} \right)^{-1} = \frac1{1 - \frac1x} = \frac{x}{x-1} = 1 + \frac1{x-1}$$ Hence, $$y' = - \frac1{(x-1)^2}$$
To proceed through you method of using chain rule, setting $u = (1-x^{-1})$, we have that $y = u^{-1}$. Hence, $$y' = -\frac{u'}{u^2}.$$ Note that $$u' = \frac1{x^2}.$$ Hence, we get that $$y' = - \frac{1/x^2}{(1-x^{-1})^2} = - \frac{1/x^2}{(1-1/x)^2} = - \frac{1/x^2}{(x-1)^2/x^2} = - \frac1{(x-1)^2}$$
The error you made was that you did not write $u^{-2}$ correctly. $\displaystyle u^{-2} = \frac1{(1-x^{-1})^2}$
@Jordan $$(1-x^{-1})^2 = (1-1/x)^2 = (x-1)^2/x^2.$$ Hence, $$\frac{1/x^2}{(x-1)^2/x^2} = \frac1{(x-1)^2}.$$ –  user17762 May 5 '12 at 18:16