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I was all set to begin Calculus 2 when I thought, "I should have a more intuitive sense of what's happening with differentials before I move on."

I want to tell you what I've learned and ask you all to help me put it together more cohesively than I have done. I'm just going to list what I know. Please point out my inconsistencies. And please forgive me if the following seems to ramble, I'm just sharing my thought process and I want it to have more rigour.

I've seen Qiaochu Yuan saying a few times that $dy$ and $dx$ are not numbers. I didn't know what he meant. So, I read elsewhere, and learned about infinitesimals. I read that an infinitesimal is a concept similar to zero, but not. It is a small, unmeasurable value. That there is no value you can multiply an infinitesimal by to get any real number. Yet, it's not zero. I read that this is an important distinction.

So, when I read $\frac{dy}{dx}$, this is not a fraction: it should be read as one symbol, and interpreted as shorthand for $\lim\limits_{\delta x \to 0}\frac{\delta y}{\delta x}$.

A problem I still have, though, and it's surprising even to me that I don't get this yet, is this: if I'm finding the gradient of a curve at a point, then I'm finding the gradient of the tangent line at that point. I do this by setting up the quotient $\frac{\delta y}{\delta x}$ where both $\delta y$ and $\delta x$ are measureable values, real numbers. as soon as I write $\lim\limits_{\delta x \to 0}$ in front of the quotient, I'm saying that the quotient coupled with the limit is not a quotient. So $\delta y$ and $\delta x$ are still real numbers, but $\frac{dy}{dx}$ is a symbol that means the tangent, the limit, or $f'(x)$.

So two things I'm not sure of at this point: it seems that if $\frac{dy}{dx}$ is a symbol in its own right, then $dy$ and $dx$ are not independent symbols. And if they are, I don't know what they mean. I will guess that $dy = \lim\limits_{\delta x \to 0}\delta y$ and $dx = \lim\limits_{\delta x \to 0}\delta x$. To say that $dx$ is infinitesimally small is to say that it is infinitely close to zero but not zero.

But, if two points are infinitely close to one another, where the distance between them is an infinitesimal, they are still two points aren't they? They can only be the same point if the distance between them is zero, and an infinitesimal is not zero. And so, how can I find the gradient of the tangent line when the tangent is really a secant?

I know this is important to understand, and at some point I thought I understood it intuitively. But my understanding has slipped and I'm here. I can tell myself "it just is" but it's not much of an argument and it's not satisfying. I want to understand everything as much as I'm capable of doing so, and take nothing on faith.

Once I understand this, I will be able to write the chain rule without multiplying by infinitesimals, then the same with the substitution rule, then I move onto Calculus 2. Until I understand this, I can't move on.

I'm guessing my problem comes from a lack of rigor in some textbooks, and that makes my overall understanding weakened. If I try to solve my own problem, I might say that $\frac{dy}{dx}$ is not, in fact, the gradient of the tangent at all. It is the gradient of the secant, but it is the value that is infinitely close to the tangent. It "approaches" the tangent. But it still comes down to this: if two values are not the same, and their difference is infinitely small, what does that mean? Sometimes they are considered the same, sometimes they are not. Which is it? If something is infinitely small, it has a value greater than zero, but it's not zero, it can't accumulate to become any real number.

It's not a number. It's a concept with the above properties and that's all there is to it. Is that the answer?

I understand the idea is to be able to get infinitely close to a value while avoiding a division by zero. And knowing that is almost enough for me to move on. But I think there's more for me to understand here, and a lot rests on it.

Thanks for reading this far, if you did. I appreciate your feedback.

Also: another question. If I have $dx$ and $du$ and they are both infinitesimals, then $1+dx = 1+du$, right? Yet, $du$ and $dx$ are not the same, right? In an integral, I can't just switch out a $dx$ for a $du$, yet they both represent the same concept.

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I think you deserve to read this answer by Arturo Madigin. math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio/… –  Patrick Da Silva Jan 4 '12 at 15:35
    
Also, I may just be asking the same question as is posted on math.stackexchange.com/questions/22725/…;. I apologise, if so. And I have read math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio/… and I understood it. I think it would be useful to completely avoid a situation where I treat infinitesimals like quotients for pure convenience. If infinitesimals are so valuable so to require creating a whole new concept, it undermines them to treat them like quotients :) –  Korgan Rivera Jan 4 '12 at 15:45
    
Treating them as quotients is the right way to think about them as long as you're not dealing with partial derivatives or more abstract notions of differentiations (such as directional derivatives or tangent spaces). It helps you remember things like $dy/dx = 1/(dx/dy)$ when $y(x)$ is an invertible real function of $x$ that is differentiable. It is not the right way to prove things though. To do that, stick with definitions you know and standard proof techniques. Dealing with infinitesimals is a pain. –  Patrick Da Silva Jan 4 '12 at 16:01
    
I'm guessing that infinitesimals are all about a convenient way to explain and understand an idea, but really they are trouble if you're being truly rigorous, and my notation is not comprehensive enough yet. They'll help me remember how to do the chain rule, but if I'm working with the chain rule like this, I'm not truly understanding what is happening. It's frustrating to know I won't understand this until later, yet I can't get to 'later' until I understand. It's like I'm not studying math anymore, I'm studying submath. –  Korgan Rivera Jan 4 '12 at 16:08
    
Thinking about infinitesimals in a rigorous way should not be seen as a necessity, that's all that I can say. –  Patrick Da Silva Jan 4 '12 at 16:41
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1 Answer

up vote 1 down vote accepted

If you are thinking about "infinitesimals" at all you're making a mistake.


Let $f$ and $g$ be differentiable functions (with derivatives $f'$ and $g'$). We are after the derivative of $f \circ g$, which, by the limit definition of derivative is:

$$\lim_{h \to 0} \tfrac{1}{h} \left[ f(g(x+h)) - f(g(x))\right]$$

the trick is to let $k = g(x+h) - g(x)$ and we can rewrite this as

$$\lim_{h \to 0} \tfrac{1}{h} \left[ f(g(x)+k) - f(g(x))\right]$$

so now another adjustment gives

$$\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \frac{ f(g(x)+k) - f(g(x))}{k}$$

and you can spot the derivative pattern twice now (one must prove that $k \to 0$), so we have

$$g'(x) f'(g(x))$$

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I'm inclined to agree, though I'm not yet qualified to. Infinitesimals are a hack! –  Korgan Rivera Jan 4 '12 at 16:25
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to clarify a little @KorganRivera, there is a beautiful rigorous theory of infinitesimal numbers, but Calc 2 is definitely not coming anywhere near it - that's what I meant. –  user16697 Jan 4 '12 at 16:29
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Mistake? Non-standard analysis has a rigorous foundation. I'm not seeing the harm if someone finds it helpful as a pedagogical tool. (I don't personally believe one method as better than the other in this regard.) en.wikipedia.org/wiki/Non-standard_analysis#Pedagogical –  dls Jan 4 '12 at 16:33
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@QED: Dear QED, "you're making a mistake" is a very categorical statement, which I would disagree with. After all, the manipulations of the quantities $dy$ and $dx$ that occur e.g. in the theory of differential equations have a meaning. It is true that naive arguments with infinitesimals are not well-founded in the standard approach to analysis that one sees in an undergraduate class, but that says something about our approach to the mathematics, not the actual mathematics itself. Regards, –  Matt E Jan 4 '12 at 17:26
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I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. –  Pete L. Clark Jan 4 '12 at 18:21
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