The following is from Stewart's 'single variable calculus, 6E' (the bold is mine)

$$\int f(g(x))g'(x)dx = \int f(u)du$$

"Notice that the Substitution Rule for integration was proved using the chain rule for differentiation. Notice also that if $u=g(x)$, then $du = g'(x)dx$, so a way to remember the Substitution Rule is to think of $dx$ and $du$ in (4) [the above equaion] as differentials."

I understand the proof that the textbook provides for the substitution rule, but it doesn't say anything about differentials. I also see that if you make the substitutions mentioned, then you have the exact same notation on both sides of the equation, but saying that $du$ can be treated as a differential seems not justified. Therefore, its not clear to me how we can say that $du = g'(x)dx$.

  • $\begingroup$ Should that be $\int f(g(x))g'(x)dx$? Also, what is Stewart's definition of a differential? It seems that Stewart is saying that the rules of differentials help someone remember the substitution rule, not (necessarily) to prove it. Perhaps that's why differentials do not appear in the proof? $\endgroup$
    – Michael E2
    Dec 29, 2012 at 3:29
  • $\begingroup$ @Michael Yes it should, thanks. I will change it. I think he is saying what you suggest, but he also says $du = g'(x)dx$ as though its a proven fact. $\endgroup$ Dec 29, 2012 at 3:35
  • $\begingroup$ To prove the equation just differentiate both sides with respect to $x$ and note that the RHS is a function of $u$ hence the chain rule applies and generates $\frac{du}{dx}$ which is precisely $g'(x)$. I assume that is the proof you understand. Notice $g'(x)dx=\frac{dg}{dx}dx$ seems to change to $du$ on the other side, it's as if the $dx$'s cancel. Javier's answer is the same point I make. $\endgroup$ Dec 29, 2012 at 4:45
  • $\begingroup$ I'm writing a story that hopes to answer your question. Un momento. $\endgroup$
    – 000
    Dec 29, 2012 at 5:33

2 Answers 2



Stewart is shorting you on the whole story. I'm going to attempt to broaden your horizons so that you can see a few flowers, pick them, and proceed on your merry way.


Back in the olden days, Leibniz considered the derivative of a function $y=f(x)$ to be the change of $y$ with respect to $x$. We get a feeling that the phrase 'with respect to' is being used much like the phrase 'in ratio to' since Leibniz would denote such a derivative with the following scribbles: $\dfrac{dy}{dx}$. Clearly, though, Leibniz considered $dy$ and $dx$ to entire quantities. By 'entire quantities', I mean that he did not think that there was multiplication occurring. That is, $dy$ is not $d$ multiplied by $y$ and $dx$ is not $d$ multiplied by $x$. Hence, we don't have that $\dfrac{dy}{dx}=\dfrac{y}{x}$ in this context because $dy$ and $dx$ are entire quantities. As you can see, this 'fraction' $\dfrac{dy}{dx}$ is already funky...

People weren't happy with the things that Leibniz was sweeping under a rug. Most importantly, he never formalized what those scribbles $dy$ and $dx$ even meant. He just said---I am oversimplifying a bit here---that those were 'infinitesimals'. People didn't like that, especially Cauchy.


Cauchy, being the guy that he is, decided he was going to set things straight. Cauchy said, "Let there be differentials $dy=f'(x)dx$ where $f'(x)$ is the derivative of $f(x)$ with respect to $x$, and $dy$ and $dx$ are simply new variables taking finite real values." The people saw that there were differentials. And it was good.

Now, why was it so good? Well, it was good because it really got rid of the notion of infinitesimals. Cauchy wasn't concerned with what $dy$ and $dx$ were, precisely. He was concerned with what $f'(x)$ was, the derivative. Unlike Leibniz, Cauchy defined the derivative as a limit of difference quotients and let the expressions $dy$ and $dx$ be defined in $dy=f'(x)dx$ as two variables which are real numbers. Hooray! No more infinitesimals.

What Stewart is Doing

I would assume that Stewart is operating in the domain Cauchy provides and not the domain Leibniz provides (though Leibniz really doesn't provide a good domain to begin with, as pointed out by Cauchy and others). So, Stewart is saying that a differential is some real variable defined by an equation of the form $dy=f'(x)dx$ where $dy$ and $dx$ are both real numbers. That's what a differential is in Stewart's context.

So, how do we arrive at the notion $du=g'(x)dx$? Purely by definition!


And here are the other flowers you missed: (1), (2), . . . :-)

  • 1
    $\begingroup$ Constructive criticism: The letters are uncomfortably large. $\endgroup$
    – Pedro
    Dec 30, 2012 at 6:01
  • $\begingroup$ @PeterTamaroff I genuinely don't know what you're referring to. The section headings? Or the fractions written using \dfrac{}{} rather than \frac{}{}? $\endgroup$
    – 000
    Dec 30, 2012 at 16:11
  • $\begingroup$ The headings. ${}{}$ $\endgroup$
    – Pedro
    Dec 30, 2012 at 17:29
  • $\begingroup$ @PeterTamaroff How is this now? :-) I chose to use ## rather than ### as ### seems equivalent to boldface normal text. $\endgroup$
    – 000
    Dec 30, 2012 at 18:01

This is kind of a short answer, but if $u=g(x)$, then $\frac{du}{dx}=g'(x)$ and so $du = g'(x)\ dx$. Remember that this is just convenient notation to help us remember the substitution rule; in this context, the differentials don't actually have any meaning.

  • $\begingroup$ Are you treating the differentiation symbol also as a fraction, $du/dx$, where the terms can be interpreted as differentials, which in turn can be interpreted as part of the integral notation? I'm sorry, but this confuses me. $\endgroup$ Dec 29, 2012 at 4:33
  • $\begingroup$ The differentials are not technically fractions, but operationally the equations makes it appear so. Formally, you can treat them as fractions for the most part in calculus with a few subtle exceptions in advanced calculus. For example, there is a place where the product of three seemingly cancelling partial derivatives give $-1$ yet naively they would seem to reduce to $1$. But, read what Javier wrote again and ponder it. $\endgroup$ Dec 29, 2012 at 4:49
  • 2
    $\begingroup$ I don't think it's a good idea to implicitly state $$[u=g(x)]\wedge\left[\frac{du}{dx}=g'(x)\right]\implies [du=g'(x)dx].$$ This ignores the fact that in the context it is being presented it is purely a definition and implies there is a multiplicative structure with differentials and derivatives. I think I'm being nitpicky here (and I apologize), but I hope this is constructive criticism. $\endgroup$
    – 000
    Dec 29, 2012 at 6:01
  • $\begingroup$ I tried to explain that this is just notation and you can't really manipulate differentials as if they were real numbers, but maybe I wasn't clear enough. I was just trying to explain the logic behind $du =g' (x)\ dx$ anyway. But I think the comments here are clear enough in explaining that the derivative is not actually a fraction. $\endgroup$
    – Javier
    Dec 29, 2012 at 13:06
  • 1
    $\begingroup$ I think I finally understand where I got confused. In the equation $du=g'(x)dx$, the term $du$ is considered to be a differential, and the truth of the equation follows directly from the definition of a differential. What confused me is that the fact of that truth has no relation to definite integrals or the proof of the substitution rule, but I assumed it must have, since it is mentioned. However, it is only mentioned in order to point out a 'trick' for remembering and applying the substitution rule. Notably, the substitution rule confused me somewhat on its own, compounding my problem. $\endgroup$ Dec 30, 2012 at 4:13

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