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I apologize for the dumb question but is the dx (or dy or similar) used in calculus a multiplier? I don't have a good grasp on what it really means or why it's there.

I'm getting to the point in integration where we are converting dx to du and it's weird for me to manipulate it like this.

An ELIF answer would be nice if that's even possible!

Douglas Marsh
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  • Most of the times it's just a symbol. It could mean multiple things, either that you integrate with respect to $x$ or you derivative with respect to $x$ (an "integral" and "partial derivative" are some key words you can search on google). It's almost never a multiplier in the usual sense. – Yanko Dec 05 '19 at 18:15
  • ELIF ? area under a curve. –  Dec 05 '19 at 18:15
  • This is a good question with many kinds of answers. Perhpaps look through posts on this site that use the word "infintesimal". These might help: https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio https://math.stackexchange.com/questions/1991575/why-cant-the-second-fundamental-theorem-of-calculus-be-proved-in-just-two-lines/1991585#1991585 – Ethan Bolker Dec 05 '19 at 18:16
  • The simplest answer is that it is just a convenient part of the notation for integrals along with the integration symbol $\int$ and its limits of integration. You could try other explanations, but they are more complicated and problematic until you have more experience and more advanced integration needs such as integrating differential forms. – Somos Dec 05 '19 at 18:29
  • Thanks. ELIF stands for "Explain it Like I'm Five". So is it more like an operator? – Douglas Marsh Dec 05 '19 at 18:30
  • Also similar: https://math.stackexchange.com/questions/200393/what-is-dx-in-integration – Hans Lundmark Dec 05 '19 at 19:36

1 Answers1

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Yes it is.

When you have a relation such as

$$y=f(x)$$

you write

$$dy=f'(x)\,dx.$$

In this relation, $dx$ is an arbitrary nonzero quantity, and $dy$ is the linear part of the variation of $y$ corresponding to $dx$; it is also an ordinary number.

Hence you can rewrite the previous equation as

$$\frac{dy}{dx}=f'(x)$$ or $$\frac{dx}{dy}=\frac1{f'(x)}$$

or

$$dx=\frac{dy}{f'(x)}.$$

In integrals,

$$dx=\frac{dx}{du}du=x'(u)\,du.$$

  • There is nothing about $dx$ that requires it to be non-zero. It is also not an arbitrary quantity. It is rather a function that inputs an increment (perhaps arbitrary and which could be zero), returns its component in the $x$-direction. – conditionalMethod Dec 05 '19 at 18:28
  • Thank you, I'm getting there. I think my grade will be better when I better understand this. I do better with reasoning rather than memorization. – Douglas Marsh Dec 05 '19 at 18:32
  • @conditionalMethod: hem, do you often divide by zero ? –  Dec 05 '19 at 18:32