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when i am learning differentiation, my lectuer tell us that the deriative $dy\over dx$ is one things, it is not the ration between dy and dx. However when i learn about integrating, sometime we need to do substitution, like integrating $\int_{0}^{1}2xdx$ when substituting $y=2x$, we can substitute $dy=2dx$, but why in this case it can be treated as 2 different terms instead of 1 term??

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marked as duplicate by Dennis Gulko, vonbrand, azimut, Asaf Karagila, rschwieb Apr 20 '13 at 12:32

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You are skipping a step, that's why. $$dy = \frac{dy}{dx}dx$$ and as $y = 2x$, $\dfrac{dy}{dx} = 2$, so $$dy = 2dx.$$

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It is good to remark that the notation $\dfrac{\mathrm dy}{\mathrm dx}$ is solely an abuse of notation. This is "justified" (rather, convenient) because in most cases, the quantity actually behaves as if it actually were the quotient (e.g. in the above identity). Once you have a rigorous definition of what $\mathrm dx$ and $\mathrm dy$ are (usually given in advanced analysis courses, e.g. when treating manifolds), you may understand that the "quotient" is simply meaningless. –  Lord_Farin Apr 20 '13 at 9:12
    
@Lord_Farin as far as i know in analysis course, they treat dx as a function. but then why the qoutient is simply meaningless? more important, what the meaning of quotient when it comes to manifold or treat dx or dy as a function? –  john Apr 20 '13 at 9:44
    
@Lord_Farin: As far as I'm concerned, $\frac{dy}{dx}$ is not an abuse of notation at all, it is the name of a function, just like $f$, $g$, or $\sin$. It is named as such precisely because $dy = \frac{dy}{dx}dx$. –  Michael Albanese Apr 20 '13 at 10:50
    
@Michael: It is abuse of notation because it is a name used to convey an intuition that is not formally applicable. It's written as a quotient, but not a quotient. Ergo abuse of notation. –  Lord_Farin Apr 20 '13 at 11:00
    
@john: It's simply not a function. In fact, I've never seen it treated as such. Things like $(\mathrm dx)^2$ and $\frac1{\mathrm dx}$ do simply not make sense. Perhaps someone else who has seen this treatment can explain it to you, but I can't. –  Lord_Farin Apr 20 '13 at 11:02

The reason is that $dy/dx$ is the limit of $\Delta y/\Delta x$ as $\Delta x \rightarrow 0$. The limit process allows us to cheat a bit and consider the derivative as the ratio of $\Delta y/\Delta x$. This serves our purpose for integration; when we write

$$dy = 2 dx$$

we mean

$$\Delta y = 2 \Delta x$$

and then we may take that limits as $\Delta x \rightarrow 0$.

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