I will be teaching Calculus 1 soon and I am trying to find some justifications for fishy arguments that are widespread out there.
In a standard Calculus 1 course, the following concepts are presented to students.
Antiderivative: A function $F$ is called an antiderivative of a function $f$ in an interval if $F'=f$ in that interval.
Indefinite integral: the family of all the antiderivatives of a function $f$ is called indefinite integral of $f$ and is denoted by $\int f(x)dx$. Having shown that the difference of any two antiderivatives of the same function is constant, if $F$ is an antiderivative of $f$, then
we write $\int f(x)dx=F(x)+C$, where $C$ is a constant.
The problem I see is that some textbooks define the differential in a very vague manner and then foster the use of the equality $dy=y'dx$ without justification.
For example, when presenting the integration by parts all starts fine with the product rule of two differentiable functions $u$ and $v$:
$$(uv)'=u'v+uv'\implies uv'=(uv)'-u'v$$
which implies that $$\int u(x)v'(x)dx=u(x)v(x)-\int u'(x)v(x)dx\quad\quad\quad (A)$$
The problem starts with the manipulation of the dummy symbols in the notation of the indefinite integral by the substitutions $dv=v'(x)dx$ and $du=u'(x)dx$ resulting in the popular formula:
$$\int udv=uv-\int vdu\quad\quad\quad (B)$$
When I look at the definition of indefinite integral, equality (A) is well-defined but (B) is not.
Into practice: Calculate $\int 2x\cos(x)dx$.
A student using (A) will write: let $u(x)=2x$ and $v'(x)=\cos(x)$. Then $u'(x)=2$
and $v(x)=\int cos(x)dx=\sin(x)$ (here undertanding that we just need 1 (any) antiderivative)
Then by (A) we have: $\int 2x\cos(x)dx=2x\sin(x)-\int 2\sin(x)dx=2x\sin(x)+2\cos(x)+C$.
When using (B) students use $u=2x$ and $dv=\cos(x)dx$. Then compute $du=2dx$ and $v=\sin(x)$, and finally replace the pieces into (B) as if they were TeX processors. I mean, the method relies on the syntax of (B), not in the definition of indefinite integral.
Question: what is the mathematical justification to accept the use of (B)? The justification should be at the level of students taking Calculus 1.
Remark: Note that substitutions of the type $dy=y'dx$ are not necessary for the substitution techniques of integration in a Calculus 1 course.
Indeed, if $F'=f$, then the chain rule shows: $$(F\circ g)'(x)=f(g(x))g'(x)$$ so by the definition of indefinite integral $$\int f(g(x))g'(x)dx=F(g(x))+C,$$ or equivalently, $$\int f(g(x))g'(x)dx=\left.\int f(u)du\right|_{u=g(x)}.$$
Update: Thanks to the answers posted, I realized that my concern was justified: (B) is (apparently) only justified after considering contents that are not part of a calculus 1 course, say, through Stieltjes integrals or differentials. Thank you for the well-presented answers and for the comments and resources presented in the comment sections.
I am well aware that it would not be good to hide (B) from my students since as it was pointed out in the comments, students will face it sooner or later and they should be prepared for it. That is why I posted this question. I think I will present and mostly use (A) during the course. I will mention (B) stating that is true but we do not have the tools to prove it and that for now it can be used as a notation-wise shortcut for (A), so they have a way to justify steps that appear in many calculus textbooks, steps that are layout without a proper justification (and you wonder why people do not understand mathematics).
Best Answer
I want to say that I strongly disagree with the view presented by OP and further I think that we are doing a disservice to students by hiding the approach B from them.
First we have the notation $$\frac{dy}{dx}=y^{\prime}$$ I think you do not disagree with this notation, even though it is not really a fraction. We can then write the expression as, $$dy=y^{\prime}dx$$ and this is just equivalent notation. Or even better to write it as $$dy=\frac{dy}{dx}dx.$$ I would present this to the students as simply notation. In this sense equation B is the same as equation A in an alternative notation.
The main point however is that expression B is much easier, especially for students to remember and use in calculation. Further, it presents significant simplifications in calculation. I find working with students that once you can get them to accept this $dy$ notation, and this may require a little practice, they make rapid progress in applications of integration. Indeed many are confused by the standard A approach that is given them and it hampers their progress. For example, this is how I write the integration of $\int x^2 \cos x dx$, $$\int x^2\cos x dx=\int x^2d(\sin x)$$ $$=x^2\sin x-\int \sin x d(x^2)$$ $$=x^2\sin x-\int 2x\sin x dx$$ $$=x^2\sin x+\int 2x d(\cos x)$$ $$=x^2\sin x+2x\cos x-\int 2\cos x dx$$ $$=x^2\sin x+2x\cos x- 2\sin x$$
This is very streamlined and it makes difficult problems easier to solve.
There is also another issue, how will you treat substitution? Will you not not write, $x=f(u)$
and so $$dx=f^{\prime}(u)du$$ Thus you will have to use the $dy$ notation on any case, so why not harmonize the two methods of integration? There is also a fact that you will have to reconcile with, students will hopefully continue in mathematics, and they will encounter the other notion, thus they should be prepared for it.
As to the question of rigor there are two options: the Stieltjes integral as noted in the previous answer (which is a good idea to indicate), or the idea of differential forms, too advanced but it is rigorous.
For me I would use the notation of the B form almost exclusively. And with the following proof. $$(uv)^{\prime}=uv^{\prime}+vu^{\prime}$$ $$uv=\int uv^{\prime}dx+\int vu^{\prime}dy$$ $$uv=\int udv+\int vdu$$ And then I would provide many exercises to promote facility with manipulation of this notation. But, of course, people have been arguing about how to teach calculus for decades.