# When $dy/dx =0$ for all $x$ in the domain, is $dx/dy$ also zero?

If $dy/dx = 0$ for all $x$ in the domain, is $dx/dy$ also zero? This seems problematic because $dy/dx$ can be thought as $0/1 = 0$ but when you reverse the upper and lower part of the fraction, the fraction is an invalid number.

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If $dy/dx=0$, then $dx/dy$ is infinite. One way to visualize this is that an inverse function $x(y)$ may be seen as a $90^{\circ}$ rotation of the original function $y(x)$. In that case, a horizontal slope becomes a vertical one by inversion.

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Do you think that we can treat $dy/dx$ as a formal fraction ike 1/4? Thanks. –  Babak S. Feb 5 '13 at 6:48
No. The nomenclature is fortunate because it makes the chain rule easy to remember, but it must be viewed as a derivative, which is the limit of a ratio of two very small quantities. In this way, the $dx$ and $dy$ may be manipulated like a fraction, but you must remember that it is a mnemonic, nothing more. –  Ron Gordon Feb 5 '13 at 6:50
+1 and thanks.. –  Babak S. Feb 5 '13 at 6:52
@rlgordonma I tend to disagree that the heuristic is nothing more than a mnemonic. It is true that it is not quite a fraction, but a limit of a ratio. The fact that it is the limit of a ratio is precicely what allows us to manipulate it like a fraction. In fact non-standard analysis allows us to view $dy/dx$ as a ratio of infinitesimals. –  Baby Dragon Feb 6 '13 at 19:01
@BabyDragon: yes, I agree with you (and I stated that comment) that we can manipulate $dy$ and $dx$ for that very reason. My point was that the "cancellation" in the chain rule is a mnemonic rather than actual cancellation. It works pretty well and makes the chain rule easy to remember, but it also misleads because one may expect to be able to substitute values for $dx$, etc., when there is no such thing. –  Ron Gordon Feb 6 '13 at 19:08

If $\frac{dy}{dx}=0$ in an open interval, then $y$ is constant in that interval, so one does not have an inverse function $x$ of $y$.

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+1 For the proper explanation. –  Jayesh Badwaik Feb 5 '13 at 7:03
+1 for the proper explanation. –  Amr Feb 5 '13 at 8:21
+1 (echoing @Amr)... –  amWhy Feb 6 '13 at 0:06

Note that $y$ is a function of $x$ and as $\displaystyle\frac{dy}{dx} = 0$, $y$ is a locally constant function. Furthermore as $\displaystyle\frac{dy}{dx}$ exists for every point in the domain, the domain must be open. As such, $y$ is not an injective function of $x$, so it can't have an inverse; in particular, $x$ is not a function of $y$ so $\displaystyle\frac{dx}{dy}$ doesn't make sense in this context.

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+1 For noting that $dx/dy$ does not make sense –  Amr Feb 5 '13 at 8:29