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I'm taking differential equations right now, and the lack of fundamental knowledge in calculus is kicking my butt.

In class, my professor has done several implicit differentiations. I realize that when taking the derivative with respect to "$x$," I have to write "$\frac{dy}{dx}$" whenever I differentiate "$y$" and nothing when it's "$x$" since it will be "$\frac{dx}{dx}$."

However, today, he differentiated the equation "$y=ux$" to get "$d y= u\, d x + x\, d u$." I'm not sure why he has decided to write $d y$ and $d x$ separately like that. I don't think he did implicit differentiation. Can someone explain what he did? Thank you very much.

EDIT: To provide more context, the professor was working on changing a homogeneous differential equation to a separable differential equation. He stated that in $f(x,y)$, all $y$ must be substituted with $ux$. Therefore, $f(x,ux)$. This prompted him to find the derivative of the equation as well, which he wrote as "$dy = u\,dx + x\,du$"

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    $\begingroup$ Ask your professor to explain differentials to you. That's what he did in the derivation you don't understand. And if you're struggling with prerequisite material this early in the semester you should really reach out to him so he can help. $\endgroup$ – Matthew Leingang Jan 8 '15 at 20:57
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    $\begingroup$ I would say that it is probably sloppy notation. I presume we're not talking differential forms here? $\endgroup$ – copper.hat Jan 8 '15 at 21:03
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    $\begingroup$ @DanSmith: I'm not sure your edit is a faithful representation of the question. This is an ordinary differential equations class, so partial derivatives are probably not being discussed. $\endgroup$ – Matthew Leingang Jan 8 '15 at 21:03
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    $\begingroup$ Editors, please revert the edit that changed $dx$ etc. to $\partial x$. I would myself but I do not have enough rep to edit. $\endgroup$ – Matthew Leingang Jan 8 '15 at 21:16
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    $\begingroup$ I just changed the $\partial x$, etc., back to $dx$. $\endgroup$ – apnorton Jan 8 '15 at 21:46
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It is simply the product rule and chain rule applied:

$$\require{cancel} y = u(x) x \implies \frac{dy}{dx} = u \frac{dx}{dx} + \frac{du}{dx}x \implies dy = u\, dx \frac{\cancel{dx}}{\cancel{dx}} + x\, \cancel{dx} \frac{du}{\cancel{dx}}.$$

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    $\begingroup$ The horror... the horror. Unless you are working with $dx$ as infinitesimals. $\endgroup$ – Asaf Karagila Jan 8 '15 at 21:04
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    $\begingroup$ @AsafKaragila It's not my horror. I'm just the messenger. $\endgroup$ – Emily Jan 8 '15 at 21:05
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    $\begingroup$ That's certainly a helpful way of looking at it, intuitively, but $dx$ is not a value, so you can't just multiply by $dx$. $\endgroup$ – Thomas Andrews Jan 8 '15 at 21:08
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    $\begingroup$ @ThomasAndrews Of course. But in a non-strict sense, you sort of can, which is the strength of the $\frac{dy}{dx}$ notation. Of course, what's being done under the hood is a different thing entirely, but I'm not the professor who decided to present it in this fashion. The idea of "bring the $dx$ over" is so ubiquitous in teaching separable ODEs that I'm not sure if digging into the analytical engine underneath would help or hinder. $\endgroup$ – Emily Jan 8 '15 at 21:11
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    $\begingroup$ @GFauxPas Not originally. The \partials were edited in. $\endgroup$ – Emily Jan 8 '15 at 21:14
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Sometimes we write $$ d y= u d x + x d u $$ and understand it to mean $$ \frac{d y}{dt}= u \frac{d x}{dt} + x \frac{d u}{dt} $$ where $x, y$ and $u$ are all functions of some variable $t$, perhaps $t$ is yet to be determined...

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$dx$ is really an enigma in calculus and differential equations. Leibniz treated it as an “infinitesimal”—a number smaller than any positive number but larger than $0$. In differential geometry it is something called a one-form. If I could drop two names at once, Raoul Bott told me a story about a class he took from J.H.C. Whitehead who, when asked, “What exactly is $dx$?” held his thumb and forefinger $\varepsilon$ apart and said “Liiiiittle bit of $x$.”

But in calculus $dx$ is called a differential. It can be treated as an independent variable with $x$. If $y$ depends on $x$, then we can add its differential variable $dy$ and relate them all with the equation $$ dy = \frac{dy}{dx} \,dx $$

Then all of the warnings in calculus that $dx$ and $dy$ aren't numbers so they can't stand alone and certainly can't be “divided” to form $\frac{dy}{dx}$ can be be swept under the rug. So Arkamis's derivation is not rigorous but it is “rigorous enough” for an undergrad intro to ODE course.

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  • $\begingroup$ I assume you are not referring to forms or infinitesimals here. How do you define the terms $dx,dy$? Personally, I think the notation is horrendous (in a calculus context) and introduces more issues than it resolves... $\endgroup$ – copper.hat Jan 8 '15 at 21:19
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    $\begingroup$ Great quotation! Congratulations (and envy) on your nearness to two of my mathematical heroes! $\endgroup$ – Rob Arthan Jan 8 '15 at 21:21
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    $\begingroup$ @copper.hat: I feel your pain. But the notation is commonplace at this level and awfully convenient. Besides, we are helping the OP understand his professor, who we certainly won't be dissuading with this thread $\endgroup$ – Matthew Leingang Jan 8 '15 at 21:23
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    $\begingroup$ @RobArthan: Raoul also told me that JHC stood for “Jesus, he's confusing.” $\endgroup$ – Matthew Leingang Jan 8 '15 at 21:23
  • $\begingroup$ You are correct, of course. But it may help the OP to realise that she/he is doing calculus without her/his seat belt fastened... $\endgroup$ – copper.hat Jan 8 '15 at 21:25
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$\partial y = u \partial x + x \partial u$ is very unusual notation here. Usually, similar notation only appears in forms like $\partial_i$, and is meant as shorthand for a particular directional derivative operator (i.e. $\partial / \partial x_i$, when you're writing everything in terms of a sequence $\{ x_i \}$ of independent variables).

If your teacher wrote $\mathrm{d} y = u \, \mathrm{d} x + x \, \mathrm{d} u$, that would make more sense, as an equation involving differentials.

It's possible your teacher was using $\partial$ to express that he wants to take a partial derivative with respect to some unspecified variable, but there's no real reason for such a thing since the relevant idea is already captured in a standard way via differentials and how they combine with directions: once you decide on a direction $v$, combining a differential $f \mathrm{d}g$ with $v$ produces the directional derivative $f \nabla_v g$.

Or in notation where you represent directions with partial derivative operators (e.g. $\partial / \partial x$ in a context where that's meaningful), combining $f \, \mathrm{d}g $ with $\partial / \partial x$ gives $f \partial g/\partial x$.

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    $\begingroup$ According to the OP, the teacher did write $\mathrm{d} y = u \, \mathrm{d} x + x \, \mathrm{d} u$. Someone else edited the $\partial$ symbols into the question, but that appears to have been a mistake. $\endgroup$ – David K Jan 8 '15 at 21:36
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A useful way to qualitatively understand the differential form your professor was using is as follows:

The total, or exact, differential dy represents a small change in the function y, which in this case depends on the "variables" u and x. (The fact that u is not actually an independent variable does not matter for the illustration given here). Therefore the total differential change in y must arise from small changes in either x, or u, or both. The contribution of each of these small changes (dx and du), to the total differential dy will depend on the specific form of the function y .. in your case it's just multiplication, $y=ux$, so the total differential is just $dy=udx + xdu$. In typical calculus style, implicit in this justification is the point that the actual magnitudes of dx and du must be sufficiently small that the values of the associated variables x and u are left essentially unchanged (and thus can be meaningfully used in the equation for the total differential).

Now, this explanation of course ignores the details of potential relationships between u and x, as in your case, where u is a function of x. That will change the details of how the differentials must be expressed mathematically, but the underlying meaning of the total differential is still the same: in order to calculate the total change in y, you must consider contributions from changes in all the "parameters" (i.e. independent variables or functions of those variables) upon which y depends.

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I'm pretty sure he was going for a variable substitution in an integral. Then you need to know how a small change in one variable affects the other variables, so you need the total differential (not the total derivative, just to be clear - different things).

This is calculated as $\mathbb dy =\frac{\partial y}{\partial x_1}\mathbb dx_1 + \cdots\,$, although when this is introduced, it could be before partial derivatives on the curriculum and the derivations are so straightforward that it is not unusual in my experience to simply do the calculations on the spot and list the partials without calling attention to them.

So: total differential to be used in variable substitution in an integral equation.

The answers from @gedgar and @arkamis are close, but neither is correct.

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