How do I interpret the "d" in derivative notation? I recently decided to declare as an applied math major and I've realized that I don't think I fully understand what derivative notation actually means. I'm kinda worried that I might have missed something and it may affect my progress in more advanced math classes like ODE which I'm taking right now.
I have always assumed that $\frac{dy}{{dx}}$ is just notation for a specific operation:
$$
\frac{dy}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{y\left( {x + h } \right) - y\left( x \right)}}{h}
$$
I understand that this is the definition of a derivative but this doesn't seem to be the full story. For example what does $dx$ mean by itself? In practice it seems like  $\frac{dy}{{dx}}$ is a ratio between two values $dy$ and $dx$ that have a mathematical meaning on their own. For example, its common to multiply both sides of an equation by $dx$. People say that $dx$ is just a small change in $x$ but what does that actually mean? One possible answer that I've just sorta assumed without thinking about it too much was
$$dx = \mathop {\lim }\limits_{a \to 0} (x + a) - x $$
But this seems wrong because it just evaluates to $dx =  \mathop {\lim }\limits_{a \to 0} a = 0$. Reading wikipedia articles about differential notation isn't helping much but it sounds like the meaning of $dx$ has changed over time?
The specific thing that prompted my question was a problem in my ODE class. Take a linear differential equation that looks something like this:
$$
\frac{dy}{{dx}} + y = x^2
$$
I know what to do in order to solve this but the solution feels hand wavey. If you multiply both sides by the integrating factor $e^{x}dx$ and then integrate both sides you get
$$
\int e^{x}dy + e^{x}ydx = \int e^{x}x^2 dx
$$
I just don't know how to interpret the RHS of the equation. Do I integrate with respect to x or y? Do I just integrate the $dx$ and $dy$ terms separately? Doing that doesn't give you $e^{x}y$ which is the actual answer.
Edit: some people are suggesting that in integral notation $dx$ is just sort of syntactical, it just shows where the end of the integration operation is. I am unsatisfied with this answer. If its just syntactical, then you would evaluate the ODE like this
$$
\int (e^{x}\frac{dy}{{dx}} + e^{x}y) dx \\
$$
If I am understanding the syntax argument correctly, you should be able to split this integral into two parts like this:
$$ \int e^{x}\frac{dy}{{dx}}dx + \int e^{x}y dx =  e^xy + e^xy = 2e^xy
$$
But this doesn't appear to be consistent with my ODE textbook, which says the actual answer is $e^xy$. They justified that by saying that the derivative of $e^yx$ is $e^{x}\frac{dy}{{dx}} + e^{x}y$ by the chain rule. This feels handwavey and leads me to believe that $ \int e^{x}\frac{dy}{{dx}}dx + \int e^{x}y dx $ is not notationally equivalent to $\int (e^{x}\frac{dy}{{dx}} + e^{x}y) dx$
 A: The notation $dx$ indeed has meant different things over the years, and means different things in different contexts even today.
Leibniz invented the notation $\frac{dy}{dx}$ in the 17th century, apparently because it fit in with a notion of calculus in which an "infinitesimal" change in $x$ resulted in an "infinitesimal" change in $y,$ and that there was a well-defined ratio between these two "infinitesimal" amounts. The notation presumably actually was that ratio.
Mathematicians ran into trouble with infinitesimals for a while.
Robinson finally figured out how to do them correctly in or around 1960, but in the meantime people had been using calculus for hundreds of years and they had figured out an epsilon-delta definition for derivatives.
With that definition, $\frac{dy}{dx}$ was (at most) a differentiation operator $\frac{d}{dx}$ applied to a function $y$,
and the $dx$ part had no meaning on its own.
People often use the notation $dx$ as part of an integral,
$\int f(x)\,dx,$ where it helps us to keep track of what the variable of integration is. But in other places people write $\int f$, which means the same thing.
The symbol $dx$ also turns up in differential forms, where it acts somewhat like a basis vector. In this context it really is a mathematical object that has a meaning standing all on its own outside of any other mathematical formula.
So the meaning of this symbol (and even whether it means anything standing on its own) depends on

*

*who is writing it and

*where they are writing it.

And on top of that, there are formal procedures (where I use the word "formal" here in the sense of "following the form of something" rather than "following strict mathematical definitions of the things named")
in which $dx$ is used as a kind of marker or particle, for example,
when we say $u = x^2$ and therefore $du = 2x\,dx$ for a U-substitution in an integral.
The notation has some remarkable mnemonic power. While technically it is not correct to cancel a $dx$ in the "denominator" with a $dx$ in the "numerator", in practice it works out remarkably well, at least in the calculus of single-variable functions.
Consider When not to treat dy/dx as a fraction in single-variable calculus? and its answers.
Since you are encountering this in applied math, where often you will learn how to solve problems without really learning why the solutions are guaranteed to be correct, I would treat the $dx$ and $dy$ as formal notations, tokens of a sort that (if you keep track of them according to the rules you have been shown)
will get your results to come out correctly. I don't think it will be very helpful to try to construct mathematical objects that $dx$ and $dy$ can represent when those symbols stand on their own.

As for the integrals at the end of the question (the concrete example of the use of $dx$ and $dy,$ it's been a very long time since I've done one of these, but I believe these are just ordinary integrals in the form
$\int f(x)\,dx$ or $\int g(y)\,dy.$
In fact the first step of the solution is really to convert the
original equation to this:
$$
\int \left(e^x \frac{dy(x)}{{dx}} + e^x y(x)\right)\, dx
 =  \int  e^x x^2\, dx.
$$
Note that I'm calling out $y$ explicitly as a function of $x$ here,
because to integrate with respect to $x$ we need to make sure we're integrating the correct function of $x.$
The left-hand side then can be restated as the sum of two integrals:
$$
\int \left(e^x\frac{dy(x)}{{dx}}\right)\, dx + \int  e^x y(x)  \,dx.
$$
For the first of these integrals, it turns out that
$$
\int \left(e^x\frac{dy(x)}{{dx}}\right)\, dx
= \int e^{x(y)} \, dy,
$$
which looks like the $dx$ terms canceled (and that is a good way to remember how this works), but it's a bit more subtle than that and is actually something that someone at some time had to prove (in a more general form) so that you can use it here.
One thing that makes this more subtle than simple cancellation of $dx$ is that the $x$ in $e^x$, which used to be the variable of integration, is now a function of the variable of integration.
If you forget that, you get the wrong answer.
Notice that
$$ \int  e^x y  \,dx \neq e^x y. $$
By writing $e^x y$ in the integrand and forgetting that $y$ is actually a function of $x$, you might fall into the trap of treating $y$ as a constant,
which would lead you to the incorrect result $e^x y$.
So here's what we actually have:
$$
\int e^{x(y)} \, dy + \int  e^x y(x)  \,dx =  \int  e^x x^2\, dx.
$$
The notation $\int e^x \, dy +   e^x y \,dx$ is just a shortened way of writing the left-hand side of this equation.
I consider the shortened notation a bit non-standard and might even
call it an "abuse of notation"--something that is convenient to write
but is actually not strictly correct notationally.
(But I am prepared to be corrected on this point: it may be that if you have a proper definition of the differentials $dx$ and $dy$ and integration over an integrand in which these differentials appear, the shortened notation is just as valid and correct as the ordinary single-variable-function notation.)
In any case, what we have at the end is just a few ordinary integrals of single-variable functions (once you correctly identify the functions).
I think you can solve it by applying integration by parts a few times;
if I'm right, there's a cancellation trick which lets you get away with leaving one of the integrals as is. (You never actually have to evaluate it.)
A: $$\int \left(e^{x}\frac{dy}{dx} + e^{x}y \right) dx =\int\left(e^{x}\frac{d}{dx}y + y\frac{d}{dx}e^{x} \right)dx=\int\frac{d}{dx} \left(e^{x}y \right)dx=e^{x}y +C$$
As to $dx$ itself under integral, then leaving historical reasons apart, we can say, that in measure theory in right of $d$ stands measure, so, formally it is separator between integrand and measure. In classical integrals its Riemmann or Lebesgue one. For example in Riemann–Stieltjes integral we have $dg(x)$ with respect to some $g$ function.
