I am reading in a book about differentiating, but I am confused with one of the steps he takes. We start with:

$$\begin{align} y &= x^{2} \\ y + \mathrm{d}y &= (x + \mathrm{d}x)^2 \\ y + \mathrm{d}y &= x^2 + x\mathrm{d}x + x\mathrm{d}x + (\mathrm{d}x^2) \end{align}$$

Now the author simplifies this to:

$$y + dy = x^2 + 2x\mathrm{d}x + (\mathrm{d}x^2)$$

I dislike how the middle term is simplified to $2x\mathrm{d}x$ instead of $2(x\mathrm{d}x)$, as I feel like it is more intuitive on what is going. As in, $2$ of the term $x\mathrm{d}x$, instead of $2x\mathrm{d}x$. But I fear writing it as $2(x\mathrm{d}x)$ may result in an incorrect distributive property.

Next, he omits the $(\mathrm{d}x^2)$: $y + \mathrm{d}y = x^2 + 2x \mathrm{d}x$.

Subtract the original $y = x^2.$:

$$\mathrm{d}y = 2x \mathrm{d}x.$$

Now here is where I get confused:

$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x.$$

How can he just divide both sides by $\mathrm{d}x$!? If the original term was $2$ of $x\mathrm{d}x$, wouldn't it have to be written out as $2x * 2\mathrm{d}x$, and thus divide both sides by $2\mathrm{d}x$ instead?

I think the root of my confusion is how to properly simplify:

$x\mathrm{d}x + x\mathrm{d}x$.

I trust that he is right, but I am looking for an explanation of why his simplification can work, and why $2(x\mathrm{d}x)$ would be incorrect.

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    $\begingroup$ Thank you all for your answers! I have read each carefully and they do build upon each other. It seems I originally suffered from a associative/distributive confusion, but I am glad I asked, since I learned more than I expected. $\endgroup$
    – user22979
    Commented Jul 11, 2013 at 20:35
  • $\begingroup$ Multiplication in associative; you don't gain anything by changing $2xdx$ to $2(xdx)$. $\endgroup$
    – Kaz
    Commented Jul 11, 2013 at 21:43
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    $\begingroup$ Nitpick: the square of $dx$ is not $(dx^2)$ but $(dx)^2$. $\endgroup$
    – Kaz
    Commented Jul 11, 2013 at 23:14
  • $\begingroup$ If the original term was $2$ of $x\mathrm{d}x$, wouldn't it have to be written out as $2x * 2\mathrm{d}x$ Why? If instead of $x\mathrm{d}x$ it was $x c$ should it end up as $2x * 2cx$ instead of $2cx$? $\endgroup$
    – jimjim
    Commented Jul 12, 2013 at 3:48
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    $\begingroup$ If you don't understand algebra you aren't ready for calculus. $\endgroup$
    – jwg
    Commented Jul 12, 2013 at 9:35

4 Answers 4


Your question is a good example of what happens when people work with infinitesimals outside of non-standard analysis: a lot of confusion. Look, the notion of $dx$ being a very tiny $x$ (less than any real number, and yet nonzero) is not precise, and it's not possible to define it correctly in standard analysis. There are some people that try defining $dx$ as $\Delta x$ when $\Delta x$ goes to zero, but this is zero by the definition of limit, so this is just garbage.

Many people ask: "why should we care if it is rigorous or not?", and well, it's just because when working with something rigorous the chance of confusion is much less than with something that is not even defined.

In the rigorous framework, we let $f: \Bbb R \to \Bbb R$ be given by $f(x)=x^2$, then by the definition of the derivative we have:

$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to0}\frac{(a+h)^2-a^2}{h}=\lim_{h\to 0}\frac{a^2+2ah+h^2-a^2}{h}$$

Now we simply reorganize the last expression, getting:

$$f'(a)=\lim_{h\to 0}\frac{(2a+h)h}{h}=\lim_{h\to0}2a+h=2a$$

So this limit exists for every $a \in \Bbb R$ and thus $f$ is differentiable with derivative $f'(x)=2x$ at every $x \in \Bbb R$.

So my suggestion is that you abandon this "intuitive" notion of infinitesimals and move to the rigorous standard analysis. You can pick Spivak's Calculus book: it's a very good book, even for self-study, and it'll show you how to deal with all of these things in a rigorous and straightforward way.

  • 1
    $\begingroup$ Thank you for the well written answer. Admittedly, I am unfamiliar with some of this notation, but your last paragraph summarizes what I got most out of your answer. I am currently reading from "Calculus Made Easy" by Silvanus, but I have heard a lot about Spivak's book. I think I will go with your suggestion and opt for Spivak's more formal book. The purpose of my study was to prepare me for an upcoming Calculus class, but I might as well learn correctly! $\endgroup$
    – user22979
    Commented Jul 11, 2013 at 20:31
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    $\begingroup$ The prevalence of this 'proof by abuse of notation' nonsense has always been baffling to me. How can a mathematician justify dropping all rigor in favor of some logically inconsistent handwaving? $\endgroup$
    – Joren
    Commented Jul 12, 2013 at 11:52
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    $\begingroup$ Unfortunately we can see this infinitesimal notation in physics and it is indeed very confusing to me often times, because I like the formal standard analysis. Is there a good document like "infinitesimal notation for dummies who like formal analysis"? $\endgroup$
    – Csaba Toth
    Commented Jul 12, 2013 at 15:11
  • $\begingroup$ Yes @CsabaToth, this idea of infinitesimals are heavily used in Physics. I don't know who teaches it in a good way, every single try ends being loose and confusing. My personal experience led me to try approaches to Physics that avoid infinitesimals like using differential forms and so on. If you know analysis on manifolds, then you can try Spivak's Physics book volume 1, it's a very interesting way of looking at Physics. $\endgroup$
    – Gold
    Commented Jul 12, 2013 at 16:42
  • $\begingroup$ @user1620696 Actually there's one field where I may benefit from the infinitesimal notation: if I write a physical simulation (or a function visualization) software code, dx could be the step "infinitesimal" time of your simulation (or visualization code). Of course because it won't really be arbitrarily small, you'll introduce and accumulate errors, but that's the way it is. Maybe that's why we see it in physics. If I'm lucky I can approach the visualization in symbolic way (closed form) too, then there's no accumulated error. $\endgroup$
    – Csaba Toth
    Commented Jul 12, 2013 at 19:14

Multiplication is associative, just as is addition:

That is, with addition, we know that $a + (b + c) = (a + b) + c = a + b + c$.

(In other words, parentheses can be omitted without causing any ambiguity).

The same is true with multiplication:

$$2\cdot (a\cdot b) = (2\cdot a)\cdot b = 2\cdot a \cdot b$$

and with multiplication, we often simply "juxtapose" the terms, omitting "$\cdot$" or "$\times$" to get $2 \cdot a \cdot b = 2ab$.

Now in your question, you are asking about simplifying: $$x\mbox{d}x + x\mbox{d}x\tag{1}$$

Here we can use the distributive property of multiplication over addition:

$$ab + ab = (a + a)b = (2a)b = 2ab$$

So, applying this to $(1)$: $$x\mbox{d}x + x\mbox{d}x = (x + x)\mbox{d}x = (2x)\mbox{d}x = 2x\mbox{d}x\tag{2}$$

  • $\begingroup$ Yes, @Amzoti: a silly typo!! Thanks for noticing! $\endgroup$
    – amWhy
    Commented Jul 12, 2013 at 1:09
  • $\begingroup$ @Amzoti: on some days, it seems, I can't "see" past my nose! ;-) $\endgroup$
    – amWhy
    Commented Jul 12, 2013 at 1:11
  • $\begingroup$ no worries, I sometimes have to make up to ten edits for silly (read embarrassing) grammar errors, which bother me as much as math errors! :-) $\endgroup$
    – Amzoti
    Commented Jul 12, 2013 at 1:12

$$2(ab) \neq 2a\cdot 2b = 4ab$$

$$2(ab) = 2a(b) = a(2b)$$

Either you take the 2 with the $x$, or with the $dx$.

[The above is assumed to be over reals]

  • $\begingroup$ Oh I see! That is about as succinct as I could have hoped for. Thanks! $\endgroup$
    – user22979
    Commented Jul 11, 2013 at 19:33
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    $\begingroup$ So depending how you distribute, you could have 2b, or not 2b? $\endgroup$ Commented Jul 11, 2013 at 21:05
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    $\begingroup$ @MartianInvader That is the question. $\endgroup$
    – Emily
    Commented Jul 11, 2013 at 21:50

"If the original term was $2$ of $x \, dx$, wouldn't it have to be written out as $2x\cdot 2 \,dx$, and thus divide both sides by $2 \, dx$ instead?"

Don't let the infinitesimal stuff throw you for a loop. I know the pure math guys are going to tear their hair out when I say this, but at your stage of the game you can just treat dx like a number. Specifically, where you have $2(xdx)$, substitute 1 for $x$ and $0.1$ for $dx$. Then you have $2(1*0.1)$. Work it out. Does this give you $2\times1\times2\times0.1=0.4$? You probably see that it doesn't. It gives you $2\times 1\times 0.1=0.2$. In the same way, $2(xdx)=2\times x \times dx = 2xdx$.

My answer is basically the same as amWhy's earlier answer, but I thought a concrete example would be helpful.

  • $\begingroup$ I don't have much hair to tear (anymore), but your answer is well put, so here is an upvote. $\endgroup$
    – imranfat
    Commented Jul 11, 2013 at 22:51
  • $\begingroup$ "at your stage of the game you can just treat dx like a number" - at some point the (dx)^2 just got omitted... $\endgroup$
    – Csaba Toth
    Commented Jul 12, 2013 at 15:33
  • $\begingroup$ The $(dx)^2$ disappearing is because the author was using the intuition behind infinitesimal, real numbers don't disappear when we square then. My personal experience with all of this makes me believe it's wiser to do things right from beginning. My first approach to calculus was with informal infinitesimals and it was kind of a pain to move to formal standard analysis. $\endgroup$
    – Gold
    Commented Jul 12, 2013 at 19:42
  • $\begingroup$ @user1620696 I haven't read the book, but does (dx)^2 get omitted because infinitesimal on the square is so small that it is negligible (=intuition)? But that's when I say you cannot treat them like numbers, since in this case it just disappears. $\endgroup$
    – Csaba Toth
    Commented Jul 12, 2013 at 20:26
  • $\begingroup$ @CsabaToth, I didn't disagree in what you said, this is really intuition based, it disappeared because of the reason you said. That's why I'm against this "intuitive framework" to approach calculus, because people work with undefined things. I was just using your example to say to the author of this answer why I think it's not good to approach calculus this way, even for beginners. $\endgroup$
    – Gold
    Commented Jul 12, 2013 at 20:32

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