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I'm new to calculus and I find it really confusing why we just ignore the dx at the end.
For example, when working on derivation of $x^2$, at the last step, we're left with

$f'(x)= 2x + dx$

But I've heard people in videos say: "Since $dx$ is super super super small, we can safely ignore it. But just ignoring it bugs me. If we choose to ignore it, should it not actually look like this:

$f'(x) \approx 2x $

Thanks for your time. Cheers :)

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    $\begingroup$ Uh, I'd not heed the "super super small" talk. We could say $f'(x) = 2x + o(x)$ as $x\to a$ or something like that. I've never seen something like $f'(x) = 2x+dx$. $\endgroup$ – Alvin Lepik Mar 29 at 11:15
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    $\begingroup$ Does this answer your question? Why is $d(x^2)= 2xdx$? $\endgroup$ – Aatmaj Mar 29 at 11:40
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    $\begingroup$ We actually ignore nothing $\endgroup$ – Aditya Dwivedi Mar 29 at 11:46
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    $\begingroup$ In the standard defintion of differentials, $dx$ is not super super small. On the opposite, it is an arbitrary number. $\endgroup$ – user65203 Mar 29 at 13:05
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    $\begingroup$ Does your calculus textbook teach you the notion of limits before getting on to derivatives? If not then time to replace the book. Another alternative is to learn calculus using infinitesimals via Keisler's book. $\endgroup$ – Paramanand Singh Mar 29 at 14:49
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The definition of $f'(x)$ involves taking a limit:

$$ f'(x) = \lim_{\Delta x\to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}. $$

For $f(x)=x^2$, you end up with $$f'(x)=\lim_{\Delta x\to 0} (2x + \Delta x)$$ and in the limit the second summand vanishes. It is not ignored, it becomes zero in the limit, which the derivative is:

$$f'(x)=\lim_{\Delta x\to 0} (2x + \Delta x) = 2x + 0 = 2x.$$

All the "$=$"-signs in this equation are actual equalities, there are no approximations done here.

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    $\begingroup$ +1 for the remark at end. No approximations are done here. Someone had to point this out!! $\endgroup$ – Paramanand Singh Mar 29 at 14:51
  • $\begingroup$ Hi, thanks for the answer! When you say and in the limit the second summand vanishes. It is not ignored, it becomes zero in the limit, which the derivative is: It was my assumption that Δx never really becomes 0. Does it actually = 0 when we take the limit? Perhaps I need to look more into limits. Thanks again. $\endgroup$ – wxyash Mar 30 at 10:24
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    $\begingroup$ @wxyash No, $\Delta x$ itself never becomes zero in this calculation, but the limit $\lim_{\Delta x\to 0} \Delta x$ does. Just like $\lim_{x\to\infty} \frac{1}{x} = 0$ without $\frac 1 x$ ever becoming zero. Have a look at the definition and a few examples of simple limits, I'm sure that will help! $\endgroup$ – Christoph Mar 30 at 15:10
  • $\begingroup$ @Christoph, Understood. Thanks a lot! $\endgroup$ – wxyash Mar 31 at 17:21
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$\newcommand{\D}{\mathit{\Delta}}$

You are starting from $$ \frac{\D y}{\D x}=\frac{f(x+\D x)-f(x)}{\D x}=2x+\D x $$ The derivative is the limit for $\D x\to 0$, where only $f'(x)=2x$ remains. In other words, the derivative is the slope of the tangent to $f$ in $x$.

You have to make clear when you work with infinitesimals and when with symbolic expressions. Then in non-standard terms, the derivative is the standard part of the difference quotient, that is, from $f'(x)\approx 2x+dx$ with $dx\approx 0$ it follows again that $f'(x)=2x$.

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  • $\begingroup$ You said it correctly, thanks $\endgroup$ – Aderinsola Joshua Mar 29 at 11:40
  • $\begingroup$ Thank you very much for the answer. So upon taking the limit, Δx actually becomes 0 right? It confuses me a little bit because my assumption was that Δx is teeny-tiny but never 0. Perhaps I need to look more into limits. Thanks again! $\endgroup$ – wxyash Mar 30 at 10:29
  • $\begingroup$ You have to be very careful, at least while learning the related concepts, to separate limes and infinitesimals. In the end one can translate it almost mechanically, $a_n\to a$ is the same as $a_N\approx a\forall N\approx\infty$, etc. However, not being careful leads into traps as Newton was confronted with by Berkeley about his version of infinitesimals or ghostly quantities that were non-zero numbers, processes going to zero and being zero at the same time, or at least in the same formula read at different times. $\endgroup$ – Lutz Lehmann Mar 30 at 10:45
  • $\begingroup$ Noted. Thank you very much! $\endgroup$ – wxyash Mar 30 at 10:53
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It boils down to the precise meaning of $ f'(x) $, which may go beyond the stage you have reached so far.

Strictly $f'(x)$ is the limit of $\big(f(x+\delta x)-f(x)\big)/\delta x$ as $\delta x$ approaches zero. There is also a strict definition of what limit means: given any expression $E(\delta x) $ with $\delta x$ in it, it has the limit $\ell$ as $\delta x$ tends to zero if for every small positive number $\epsilon$ you can make $$ |E(\delta x) - \ell| \leqslant \epsilon $$ for all sufficiently small but not zero $\delta x$.

Written even more precisely, if for every $\epsilon >0$ there exists a number $\Delta > 0$ (depending on $\epsilon$) such that $$ |E(\delta x) - \ell| \leqslant \epsilon $$ for all $0<|\delta x| \leqslant \Delta$, then $E(\delta x)$ has limit $\ell$ as $\delta x$ tends to zero. The strict inequalities are important. The notion is that $E(\delta x)$ gets closer and closer to $\ell$ as $\delta x$ gets smaller (but not zero). But beware. The limit does not always exist.

All this is usually written as $$\lim_{\delta x \to 0} E(\delta x) =\ell$$ or $E(\delta x) \to \ell$ as $\delta x \to 0$.

Mathematically there then follows a process of proving all kinds of different properties of limits, such as the sum or product of two limits is the limit of the sum or product and so on.

In you example $f(x) = x^2$, so you would argue: take any $\epsilon > 0$. Then for any $\delta x$, $$ \frac{(x+\delta x)^2 - x^2}{\delta x} = 2x + \delta x $$ which leads to $$ \left|\frac{f(x+\delta x) - f(x)}{\delta x} - 2x\right| = |\delta x|.$$ Now, if you choose $\Delta = \epsilon$, for every $\delta x$ with $|\delta x| < \Delta$ you have the inequality you want and you can say $$ f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x)-f(x)}{\delta x} = 2x.$$ With practice and using the many of properties of limits this gets easier.

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  • $\begingroup$ Thanks, @Wa Don! I will look into limits. Do let me know if you have any sources that can help me understand limits more precisely. $\endgroup$ – wxyash Mar 31 at 17:26
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$f'(x)=2x+dx$ is technically invalid. You cannot add functions and differentials. A correct expression is $df=2x\,dx$, but that does not help.

Using small increments, you can indeed write $$0<\Delta x\ll x\implies f'(x)\approx\frac{(x+\Delta x)^2-x^2}{\Delta x}=2x+\Delta x.$$

Then,

$$f'(x)=\lim_{\Delta x\to0}(2x+\Delta x)=2x.$$

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  • $\begingroup$ Especially for a calculus novice, it might be worth making it explicitly clear that $df=2x\,dx$ is a correct but different expression. $\endgroup$ – David Z Mar 29 at 20:43
  • $\begingroup$ @DavidZ, Noted. Thank you! $\endgroup$ – wxyash Mar 30 at 10:33
  • $\begingroup$ @Yves. Thank you very much for the answer. I guess limits is what I find a little bit confusing and I'd need to look into them. For Δx to vanish when we take the limit, it has to become 0 right. This is what I find confusing because I assumed Δx never really becomes 0, even when you take a limit. Thanks a lot again! $\endgroup$ – wxyash Mar 30 at 10:35
  • $\begingroup$ @wxyash: this is true, $\Delta x$ may not be zero, otherwise the ratio used to determine the derivative is undefined. Limits are there to tell you what value to supply by continuity where an expression is undefined. $\endgroup$ – user65203 Mar 30 at 11:41
  • $\begingroup$ @YvesDaoust, I'm going to study limits in greater detail. Thank you very much. $\endgroup$ – wxyash Mar 31 at 17:25
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Too long to be a comment, but probably not satisfactory as an answer.

$dx$ is notation and it stands for 'differential with respect to $x$'. You would never write $f'(x) = 2x + dx$, because on the LHS, there's a derivative and on the right .. I don't even know what that is tbh. If it was something like $(2x+a)dx$ then it would be a differential. The symbol $dx$ does not appear by itself. Either way, the equality $f'(x) = 2x+dx$ is suspect.


On another angle, by definition $$ \lim _{h\to 0} \frac{f(x+h)-f(x)}{h} =: f'(x) $$ if it exists. So, we can approximate $f(x+h)$ linearly, writing $f(x+\Delta x) \approx f'(x)\Delta x + f(x)$, where $\Delta x$ is 'super super small'. But $\Delta x$ and $dx$ are completely different beast. $\Delta x$ is some fixed number, whereas $dx$ is not a number at all.

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  • $\begingroup$ noted. Thank you very much. I may have confused myself with these notations. This helped me clear it. Thanks again! $\endgroup$ – wxyash Mar 30 at 10:40
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You are basically correct, and the proper symbol is actually $\simeq$. If I remember correctly, Liebniz chose a sign that was not strictly equal to make this point abundantly clear. However, there are indeed definitions of the derivative such that $=$ is proper, though I do think $\simeq$ is clearer.

Basically, in nonstandard analysis, the derivative is wrapped with the standard part function, std(). So, f'(x) is actually $\mathrm{std}(2x + dx)$. The standard part function is like a "rounding" function, which rounds to the nearest real number. So, if you are using this definition of the derivative then: $$f'(x) = \mathrm{std}(2x + dx) = 2x$$ In any case, even without the standard part function, since $dx$ is infinitesimally small, you can use $=$ without causing any major problems, and this is easier for most people, so that is what they do.

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  • $\begingroup$ No, in non-standard analysis you need to take the standard part to get rid of infinitesimals. "$\mathrm dx=0$" is false in non-standard analysis when $\mathrm dx$ is an infinitesimal. However, this question most certainly is not about non-standard analysis at all and thus this interpretation is of no use. Also the proper symbol in both standard and non-standard analysis is "$=$" in "$f'(x)=2x$". There is no approximation and using "$\simeq$" would be wrong. $\endgroup$ – Christoph Mar 29 at 11:51
  • $\begingroup$ I didn't say $dx = 0$. I also didn't say $\simeq$ was the correct symbol in non-standard analysis. I'm not sure where you got either of those. The question was about calculus generally, and nonstandard analysis is the most straightforward description of calculus that makes actual use of $dx$ as an independent entity. $\endgroup$ – johnnyb Mar 29 at 12:08
  • $\begingroup$ You said that "$\simeq$" would be the proper symbol. You also said that you can write "$2x+\mathrm dx=2x$" without the standard part function and with "$=$" and this would be what most people do. It is not. And the major problem caused by this is that it implies $\mathrm dx=0$, which is false. $\endgroup$ – Christoph Mar 29 at 15:11
  • $\begingroup$ No, it isn't. I said very explicitly that $dx$ is infinitely small and that leaving it off doesn't create major problems. This is not the same as saying $dx = 0$, it is saying "$dx$ is very similar to zero and generally treating it as such doesn't cause too many problems". That is indeed true, and IS NOT the same as saying unequivocally that $dx = 0$. $\endgroup$ – johnnyb Mar 29 at 17:02
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    $\begingroup$ Thanks for the discussion guys. I believe its something that I need to look into further. But I feel good to be surrounded by helpful and really really smart people of this community. Thanks for your reply @johnnyb $\endgroup$ – wxyash Mar 30 at 10:42
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Warning: I'm about to use the terms "secant" and "tangent" with geometric meanings I'll link to, and not their meanings in trigonometry. I didn't choose the terminology.

For $y=x^2$ the secant from $(x,\,y)$ to a point $(x+\Delta x,\,y+\Delta y)$ with $\Delta x\ne0$ has gradient $\frac{\Delta y}{\Delta x}=2x+\Delta x$. The tangent at $(x,\,y)$ is the line through it whose gradient is the $\Delta x\to0$ limit $2x$ of the secant's gradient $2x+\Delta x$. We are not ignoring the $\Delta x$ term in the secant's gradient; we are interested in the tangent's gradient, which is denoted $\frac{dy}{dx}$. This is not literally a ratio of two quantities $dy,\,dx$, but has this notation because it is a limit of a ratio.

As has been discussed, the fact that $\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=\frac{dy}{dx}$ is finite implies the small-$\Delta x$ little-$o$ result $\Delta y\in\frac{dy}{dx}\Delta x+o(\Delta x)$.

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This long winded comment is not an answer.

There is an underlying problem. New Calculus students are often taught problem solving techniques without being taught the theory. Generally, this is a good approach, because the theory can be difficult for new students to grapple with.

However, it often leads to situations very similar to this one, where the student asks :
Why is this formula correct?

A case can be made that the query specific responses that your query generated are band-aids on the underlying issue. This situation will probably repeat itself. You will find yourself using formulas or problem solving techniques without being absolutely sure that the formulas/techniques are valid.

My recommendation:

  • Keep a careful list of such questions regarding situations where you were unclear about the validity of an approach.

  • At the end of the course, when you have completely mastered the problem solving techniques, there will be many formulas/techniques/theorems that your class directed you to accept, without question.

  • After completing the course, your math education/sophistication/intuition around Real Analysis (aka Calculus) will be strong enough that you can then find a book that focuses much more deeply on theory.

Alternatively, if you feel that you can immediately (i.e. simultaneously) attack the theory, then simply immediately try to find the right supplemental text.

Generic tips for finding the right Real Analysis text:

  • The book must be customized to your abilities. Typically, a book that is good for one student won't be best for another student.

  • With the previous point in mind, Amazon reviews on a book often provide a good idea on whether the book is right for you.

  • Also with the first point above in mind, search on mathSE for queries via the string:
    [soft-question] [reference-request] [calculus]
    or perhaps
    [soft-question] [reference-request] [real-analysis]

    The brackets indicate that you are looking for mathSE queries with the corresponding tags. The soft-question tag refers to
    questions whose answers can't be objectively evaluated as correct or incorrect, but which are still relevant to this site.

    Edit
    I totally overlooked the [book-recommendation] tag.

  • You generally want books that offer many, many problems for you to solve. You typically learn more by working through the problems, which apply the theory, than by reading about the theory.

  • You might have to go through 5 or more books before you find just the right book. One approach is to find and download free pdf files of the corresponding candidate books. Then, when you've found the right one, buy it, since it will be easier to learn from the physical copy of the book.

There will undoubtedly be problems in any math book that you are attacking that you can't solve. I typically set a time limit of (30 minutes to 90 minutes). If I can't at least find a promising line of attack after spending the appropriate time on the problem, then mathSE becomes an extremely convenient resource. Structuring your mathSE query to influence reviewers to help you is tricky. I suggest seeing this mathSE article for query guidelines.

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    $\begingroup$ Really helpful tips. I'll make sure to pin it in my mental bulletin board, Thanks a lot! $\endgroup$ – wxyash Mar 30 at 10:43
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In most cases, what you are doing is simply not the way you define the limit; at no point does a "$dx$" term appear. I am going to present an approach to derivatives via nonstandard analysis, which does involve a $2x+dx$ expression at some point. It's not the usual approach to derivatives, though it is equivalent to the ordinary definition.


From the point of view of an infinitesimal calculation, we get $$ \frac{f(x + dx) - f(x)}{dx} = \frac{(x + dx)^2 - x^2}{dx} = 2x + dx. $$ This is true for any values of $x$ and $dx$. In particular, if we're working over the hyperreals, it is true for an infinitesimal $dx$.

In nonstandard analysis, we define the derivative as follows. If the expression $$ \frac{f(x+dx) - f(x)}{dx} $$ has the same standard part for every infinitesimal $dx$, then $f'(x)$ is defined as the standard part of that expression. In other words, we specifically have an "ignore the $dx$" step in this definition of $f'(x)$.

In our case, this expression is $2x + dx$, which has standard part $2x$. This is indeed the same for every value of $dx$, so we define $f'(x) = 2x$.

To appreciate how this works, let me give two examples where this does not work:

  • Suppose we try to take the derivative of $\sqrt{x}$ at $x=0$. Then we get an expression $\frac{\sqrt{0 + dx} - \sqrt0}{dx} = \frac1{\sqrt dx}$, which has no standard part: the standard part is only defined for finite hyperreals, and when $dx$ is infinitesimal, $\sqrt{dx}$ is also infinitesimal, and $\frac1{\sqrt{dx}}$ is infinite. No derivative exists.
  • Suppose we try to take the derivative of $|x|$ at $x=0$. Then we get an expression $\frac{|0 + dx| - |0|}{dx} = \frac{|dx|}{dx}$. This simplifies to $1$ when $dx > 0$, and to $-1$ when $dx < 0$, so the standard part exists but is not independent of $dx$. Again, no derivative exists.
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