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I know that

$\dfrac{d(xy)}{dx} = y$

but what does

$\dfrac{dx}{d(xy)} =\, ?$

I know this is an odd equation, but it comes from some ugly change of variables and I am stuck with it.

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  • $\begingroup$ Try the definition of a derivative as a limit. It will seem clearer for you. $\endgroup$ – guaraqe Mar 27 '13 at 16:55
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Let $z=xy$. Then $d(x)/d(xy) = d(xy y^{-1})/d(xy) = d(zy^{-1})/d(z)$.

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Yet another answer: $ \dfrac{du}{dv} \dfrac{dv}{du} = 1$ by chain rule. This means the answer is $\dfrac{1}{y}$.

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  • $\begingroup$ Your last setence assumes $y$ is constant as a function of $x$. But your first setence should still be true even if $y$ depends on $x$. $\endgroup$ – Michael Hardy Mar 27 '13 at 17:12
  • $\begingroup$ Oh, I assumed it was given he gave us the derivative of $xy$ with respect to $x$ as just $y$ and not $y + x \frac{dy}{dx}$ $\endgroup$ – muzzlator Mar 27 '13 at 17:16
  • $\begingroup$ Actually, he did say that, but I'd have generalized it anyway. I think it's worth knowing that your first sentence is true somewhat generally. $\endgroup$ – Michael Hardy Mar 27 '13 at 17:17
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In order to conclude that $\frac{d(xy)}{dx}$, one needs to know that $\frac{dy}{dx}=0$, which would make $y$ a constant. I note this because typically $y$ refers to a variable that might change with $x$. In this case,the product rule tells us that $$\frac{d(xy)}{dx}=y\frac{dx}{dx}+x\frac{dy}{dx}=$$ $$y+x\frac{dy}{dx}$$. Then $$\frac{dx}{d(xy)}=\frac{1}{y+x\frac{dy}{dx}}.$$ Under the assumption that $\frac{dy}{dx}=0$, or alternatively that $y$ is a constant, the other answers are correct. I did want to give a warning that $y$ typically is not a constant however.

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Or, in an easier way, if x and y are independent (and you know that because d(xy)/dx=y), you can treat y as a constant. so the answer is: 1/y

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  • $\begingroup$ It help readability to write answers using MathJax (see FAQ). Regards $\endgroup$ – Amzoti Mar 27 '13 at 17:34

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