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I've read many question/answer threads here on SE re: justification for the algebraic manipulation of $\mathrm{d}y/\mathrm{d}x$ in the standard formulation of calculus. I worked up my own shot at a justification using the definition of the derivative and would like to get the input of others whether it holds. It doesn't involve pushforward maps, non-standard analysis (two things I don't yet know), or any other tool than simple limit algebra/laws, the definition of the derivative, and (implicitly), the chain rule that makes the substitution possible. If it does hold as a justification as far as it goes, i.e., in one-dimension, it would seem to me, at least, to be a simple and clear justification in the form of an example that could be useful to other beginners. If goes wrong somewhere please let me know for my own understanding.

Integrate $\int \frac{x}{1+x^2}\mathrm{d}x$.

We'll do this by substitution. We want to restate the original problem in terms of $u$. That means that we want to state $x$, $1+x^2$, and $\mathrm{d}x$ in terms of $u$. Let $u=1+x^2 \rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}=2x \rightarrow \mathrm{d}x=\frac{\mathrm{d}u}{2x}$. Let's justify that last step as that's where the justification (that I speak of) is required:

By definition, $\frac{\mathrm{d}u}{\mathrm{d}x}=2x$ is defined as $$ \lim_{\Delta x \to 0}\left( \frac{u(x+\Delta x)-u(x)}{\Delta x} \right)=2x $$ Recall that the limit of a quotient is equal to the quotient of the limits as long as the denominator is not equal to zero, thus the foregoing becomes $$ \frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x) \right)}{\lim_{\Delta x \to 0} \Delta x } =2x $$ Multiplying both sides by $\lim_{\Delta x \to 0}\Delta x$ and then dividing both sides by $2x$ yields $$ \frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x) \right)}{2x} =\lim_{\Delta x \to 0} \Delta x $$ which, in the limit, is equal to $$ \frac{\mathrm{d}u}{2x}=\mathrm{d}x $$ Rearranging we have $$ dx=\frac{\mathrm{d}u}{2x} $$ as desired.

Joe
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    But $\lim_{\Delta x \to 0} \Delta x = 0$ and you can't divide by $0$ ! – Tryss Feb 26 '15 at 21:52
  • But, say, we manipulate prior to taking the limit since the limit laws allow us to separate the numerator and denominator. In other words, even in the definition of the derivative, the denominator doesn't go to zero otherwise then it wouldn't be a valid definition. It is a limiting process, i.e., a relationship, it doesn't mean literally plug in zero. Your thoughts? – Joe Feb 26 '15 at 21:59
  • My thought is that $du$ and $dx$ are not numbers, so you need another formalism to manipulate them. What's $dx$ for you? It's not $\lim_{\Delta x \to 0} \Delta x$, because in the usual sense, this is equal to $0$. So how do you define it? – Tryss Feb 26 '15 at 22:03
  • Instead of answering your question just yet, is there a problem with the limit algebra used? I mean, we can separate the limit of the quotient into the quotient of the limits, correct? And, after doing so, we can carry out the algebraic manipulations to isolate $\mathrm{d}x$ on one side, correct? These are sincere questions. – Joe Feb 26 '15 at 22:19
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    No, we can't, because the limits are equal to 0. It's only when both limits exist, are finite and not equal to 0 that you can separate them – Tryss Feb 26 '15 at 22:20
  • Ok, well there you go. Thanks Tryss. That is the crux then of why this won't work and why you made your first comment. – Joe Feb 26 '15 at 22:21
  • It'd be great for someone to use the example and show the steps for a beginner like me of justifying that step using some (other) valid way using the tools of standard 1st year calculus, e.g., with the chain rule. – Joe Feb 26 '15 at 22:24

1 Answers1

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The problem is to define $dx$ and $du$... so writing $dx = \frac{du}{2x}$ is non-trivial, and it's not easy to give a rigourous meaning to both of the terms. Sure, as a mnemonic, $\frac{du}{dx} = 2x \Leftrightarrow dx = \frac{du}{2x}$ is fine. but, I repeat, what is exactly $dx$?

What's easier to show is the substitution rule directly. As $\left(f(g(x))\right)' = g'(x)f'(g(x))$ we have

$$\int_{f(a)}^{f(b)} g(x) dx = \int_{a}^{b} g(f(t))f'(t) dt$$

Indeed, let's call $G(x) = \int_0^x g(t) dt$. We have then (when g is continous) $G'(x) = g(x)$ and

$$ \int_a^b g(f(t))f'(t) dt =\int_a^b G'(f(t))f'(t) dt $$

We now use the chain rule. as $G'(f(t))f'(t) = (G(f(t)))'$, it imply

$$ \int_a^b G'(f(t))f'(t) dt = \int_a^b (G(f(t)))' dt = G(f(b))-G(f(a)) = \int_{f(a)}^{f(b)} g(x) dx $$

So the substitution rule is not hard to demonstrate. It's harder to remember, and here $dx = \frac{du}{2x}$ can be usefull as a mnemonic. If you want to give a rigourous meaning to this expression, you'll have some work to do (and definitions to create). Also remember that for a beginner, (with Riemann integration) is a notation of

$$\int_a^b f(t) dt = \lim_{n\to +\infty} \sum_{i=0}^{n-1} f(x_i)(x_{i+1}-x_i)$$

When the limit is the same for all subdivisions. We could have noted it $I(a,b,f)$, and the substituting rule would be

$$I(f(a),f(b),g) = I(a,b,f'\times(g\circ f) )$$

Here there is no $dt$

Tryss
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  • Thanks for this @Tryss. Would you be willing to update your answer to show using the chain rule in this example explicitly? I have trouble getting from the definition of the chain rule to its application here. Just for the one step highlighted in the example. – Joe Feb 26 '15 at 23:06
  • Explained a little more the part about the chain rule – Tryss Feb 26 '15 at 23:09
  • Thanks Tryss for the update. It'd still help to see how the chain rule could be used in the example explicitly. To add to the discussion, I found Brendan Cordy's explanation located here (partway down the page) for why we can manipulate $\mathrm{d}y/\mathrm{d}x$ satisfying, mainly because it mirrors the explanation that Apostol provides in his 1st Edition (at least) Calculus Vol 1. For me, learning non-std-analysis and/or differential forms would be the way to feel fully satisfied about this though I guess. – Joe Feb 26 '15 at 23:35
  • Perhaps it is wiser for me to leave it as Christian Blatter does here: be satisfied that if the result is correct, i.e., can be shown to be correct, then all that can be said is that, in principal, there is "some" justification. Not very satisfying, but it releases me from obsessing about this when I should be doing other things. – Joe Feb 26 '15 at 23:51
  • Probably the most straightforward and organized approach to justifying $dy/dx$ manipulation using the chain rule that I have found is located here. Hope this helps someone. – Joe Feb 28 '15 at 06:20
  • Section 10.3 of Jerry Shurman's Calculus of One Variable also has a nice discussion (for beginners) of antidifferentiation by forward substitution which is really just using the chain rule in the forward direction. It helped me to see the problem in terms of the chain rule only without recourse to using Leibniz notation in the fractional sense. Here is my workup of the example I presented above using Shurman's guidance. – Joe Mar 01 '15 at 04:29