Rather simple chain rule question

I have no idea what I am doing wrong, I have run through this many times and it has to be the right answer but the book gets something different that seems impossible.

I am suppose to find the derivative of $y= (1-x^{-1})^{-1}$

This is really easy, no tricks here. I use the chain rule and I get

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

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

So this all seems correct to me but is the wrong answer, and I do not know what is wrong with what I have done up to this point.

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Could you tell us what the given answer is? – Paul Mar 7 '12 at 1:04
It could be that you're supposed to simplify this to $-(x - 1)^{-2}$. – Dylan Moreland Mar 7 '12 at 1:05
That looks right to me, may just be a simplification problem? Or the book answer is wrong... – user23784 Mar 7 '12 at 1:06
$-(x-1)^{-2}$ is the answer but no matter what I do I can not force it to be that form. – user138246 Mar 7 '12 at 1:07
Minor nitpick: on the left side, you want $dy/dx$, and when you write $dx/dy$ on the right side you really want $d/dx$. – Dylan Moreland Mar 7 '12 at 1:10

You are quite right. Perhaps the "correct answer" you are given is $-\frac{1}{(x-1)^2}$, but that is equivalent to what you have as $$\begin{eqnarray} -(1-x^{-1})^{-1}(x^{-2})&=&-\frac{1}{(1-x^{-1})^2}\frac{1}{x^2}\\ &=&-\frac{1}{((1-x^{-1})(x))^2}\\ &=&-\frac{1}{(x-1)^2}. \end{eqnarray}$$ A minor quibble about notation though: $\frac{dy}{dx}$ is a function, not an operator, so when you write $\frac{dy}{dx}(1-x^{-1})$ it means "$\frac{dy}{dx}$ times $(1-x^{-1})$", which is surely not what you intended. You should instead write $\frac{d}{dx}(1-x^{-1})$, which means "the derivative of $(1-x^{-1})$ with respect to $x$".

Edit: I didn't notice you wrote $\frac{dx}{dy}$, which is totally backwards. It should read $\frac{dy}{dx}$ except where noted in my answer.

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I really do not understand Leibniz notation so why is that the case? I mean I know how to use Leibniz notation but I do not really know why. – user138246 Mar 7 '12 at 1:09
In short, $\frac{d}{dx}$ is an operator which takes a function and produces another function. $\frac{dy}{dx}$ is a convenient shorthand for $\frac{d}{dx}y(x)$, where $y(x)$ is some differentiable function of $x$. – Alex Becker Mar 7 '12 at 1:15

You made no mistake, this is the derivative. You can simplify it a bit using $a^n b^n = (ab)^n$.

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

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It looks almost OK, sort of. Well, not really, I was just being polite.

The $\frac{dx}{dy}$ in front is wrong, we are calculating $\frac{dy}{dx}$. And the second $\frac{dx}{dy}$ does not make sense either.

But, apart from a few serious misuses of notation, an answer that is not incorrect was arrived at. That answer can be simplified in various ways. Another version of a correct answer is $$-\frac{1}{\left(x-1\right)^2}.$$

The various versions are not quite equivalent, since the original function is not defined at $x=0$.

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