I've been taught about $dy/dx$ and how it can be split into $\frac{dy}{dx}=\frac{d}{dx}y$. I'm confused as to why this happens. Don't $dy$ and $dx$ both refer to infinitely small changes in their respective variables? In that case, what is different about a $dy$ that allows it to split into $d$ and $y$?
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One of the ideas behind the notation is to denote a change in $y$ wrt $x$, so too you can denote a change wrt $x$ by an analogous shorthand $\frac{d}{dx}$. – gt6989b May 05 '22 at 03:45
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What's different is that usually $y=f(x)$ so it makes sense to ask how $y$ changes with respect to $x$. There is no such thing as an infinitely small change. – John Douma May 05 '22 at 03:48
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And this is one more reason to avoid the differential notation for derivatives... – azif00 May 05 '22 at 03:49
2 Answers
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This is just a notational convention.
$\frac d{dx}y$ does not mean anything different than $\frac{dy}{dx}$. When instead of having $y$ we have some expression we want to differentiate and do not wish to introduce some new variable just for the sake of one equation, writing something like $$\frac{d\left(\frac{e^{-2x^2} - 7x}{\log_\pi(x^2-1)}\right)}{dx}$$ is not really convenient to work with. So instead we prefer to write $$\frac d{dx}\left(\frac{e^{-2x^2} - 7x}{\log_\pi(x^2-1)}\right)$$
Admittedly, there is a somewhat different concept involved in the $\frac d{dx}y$ notation as opposed to $\frac{dy}{dx}$. You can think of it as the operator $\frac d{dx}$ acting upon the function $y(x)$. But even in this case, the operator $\frac d{dx}$ is an atomic notation, not some conglomeration. It is not "something called $d$" divided by "something called $dx$". It is defined only as a whole.
Paul Sinclair
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When we prefix $\Delta$ to a variable, it implies a discrete difference: $$\Delta x = x_2-x_1$$ where $x_2$ and $x_1$ are two values that the variable $x$ can assume. In this case, these two values can have a finite difference. When the two values approach each other (as shown in the limit below), the difference approaches to zero: $$\text{as }{x_2\to x_1},\ \Delta x = 0$$
Wrong Way to Think about Differentials
However, some mathematicians interpreted $\text{as }{x_2\to x_1},\ \Delta x \neq 0$ and rather an infinitesimally small number. Unfortunately, this interpretation pertains till date and is taught in many schools and textbooks.
$$ \lim_{x_2\to x_1}\Delta x \equiv dx $$
In other words, operator $d$ is same as operator $\Delta$ under the said limit. Prefixing $d$ to any element (or variable like $x$) operates on the element in order to create an infinitesimally small (or in some approximate methods, very small) difference of values of that variable is considered.
Thus $dx$ means a very small difference on $x$ while $dy$ means a very small difference on $y$.
As pointed out in the comments as well, this interpretation is just wrong.
Less Wrong Way to Think about Differentials
Prefixing the operator $\dfrac{d}{dx}$ means evaluating a derivative of the element following the operator with respect to x. $\dfrac{d}{dx}y$ means to find the derivative of $y$ with respect to $x$. The notation $\dfrac{dy}{dx}$ is derived from the tangent-slope interpretation of the derivative, that is to take the ratio of the opposite side with the adjacent side. The discrete product (using $\Delta$) provides the slope of the secant. Consider a secant along $x$ to a function $y$:$$ \text{slope of secant} = \dfrac{\Delta y}{\Delta x}$$ To calculate the running slope of the tangent to the function:$$\boxed{\text{slope of tangent} = \lim_{\Delta x \to 0} \dfrac{\Delta y}{\Delta x} \overset{\text{old notation}}{=} \dfrac{dy}{dx}}$$
Thus, $\dfrac{d}{dx} y$ is less wrong way to use the derivative-along-$x$ operator on a function $y$ while $\dfrac{dy}{dx}$ is a notational convention hinting towards the slope interpretation of the derivative operator.
Precautionary Note
There is no algebra involved here. $y$ is not being multiplied with $\dfrac{d}{dx}$, we are simply switching notation for convenience.
ananta
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Your definitions are wrong, 1: $\lim_{\Delta x \to 0} p\Delta x= 0$. 2: $\lim_{\Delta x \to 0} \frac {\Delta y}{\Delta x}$ does not make any sense because $y=f(x)$, which you need to indicate. Even then, it is definitely NOT equal to $\frac {dy}{dx}$ – Bertrand Wittgenstein's Ghost May 05 '22 at 22:35
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3This answer is just straight up wrong. First of all, $\lim_{x_2\to x_1}\Delta x=0$, so setting $\mathrm{d}x=\lim_{x_2\to x_1}\Delta x$ literally just means that $\mathrm{d}x=0$. There is no "infinitesimal" here. The limit equals zero. – Lorago May 05 '22 at 22:38
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Thank you for pointing out the ambiguity in assigning the equal to sign to the limit. $x_2 \to x_1$ doesn't mean that $x_2 - x_1 = 0$. The limit equations have been corrected. – ananta May 06 '22 at 00:24
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@ananta doesn't change that fact that almost everything else is wrong. $\frac {dy}{dx}$ is NOT $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ – Bertrand Wittgenstein's Ghost May 06 '22 at 03:34
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1@BertrandWittgenstein'sGhost - ??? When you break down the meanings of $\Delta y$ and $\Delta x$ here, $\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}$ is exactly the common definition of $\frac{dy}{dx}$. – Paul Sinclair May 06 '22 at 11:04
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@ananta - on the other hand, replacing the equals sign with an equivalence sign in your earlier formula merely changed it from being patently false to being patently meaningless, and therefore of no value. Differentials have been a very problematic concept in mathematics, and while several rigorous definitions have be proffered in the last century or two, all have weaknesses and require significant sophistication to understand. The concept you are pushing here, is not amongst them, as it is fundamentally flawed. Best to think of them at this stage as intuitive, not rigorous math. – Paul Sinclair May 06 '22 at 11:16
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So you will not listen to me, in that case, you will listen to Apostol. Please refer to Calculus Volume 1, 2nd edition: Section 4.8 by Apostol for reference. Let me emphasize that, as I mentioned, the limit definition of differential is not quite correct, it is a notational convention based on a historical interpretation of the derivative which 'hints' towards to slope of a tangent. The better interpretation is looking at $\dfrac{d}{dx}$ as an operator that acts on y. STOP downvoting folks! – ananta May 06 '22 at 16:53
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@PaulSinclair It's not $\frac{dy}{dx}$ stop confusing derivetive of a function, defined in terms of limits and quotients, with a differential. Yes, $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=\frac{df}{dx}$, for some particular function $f$, but each $dy,dx$, is a differential, the two are not the same. The former is the derivetive of some particular function, while the latter elements of a set with cardinailty two. – Bertrand Wittgenstein's Ghost May 06 '22 at 17:55
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One thing to consider, in the former case, where we take the limit of a quotient; the quotient $dy$, $dx$ can be undefined (or $0/0$) but not in the latter case because they won't be orthogonal anymore. – Bertrand Wittgenstein's Ghost May 06 '22 at 18:03
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1@BertrandWittgenstein'sGhost - You are the one who is confused here. In every means of defining differentials, rigorously or not and including the one you are trying to describe (but are failing to do so properly), $\frac{dy}{dx}$ is always the limif of the difference quotient. In most of them, it is that limit by definition, and then $dy = \frac{dy}{dx}dx$ is used define the differential $dy$ in terms of the independent differential $dx$. There are some ways of defining diffs. where this arises as a theorem instead, but in what you are describing, it is a definition, not a theorem. – Paul Sinclair May 06 '22 at 21:11
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@PaulSinclair it's alright... If that makes you happy, still doesn't change the fact that you are wrong. – Bertrand Wittgenstein's Ghost May 06 '22 at 23:02
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So I understand that $dy$ and $dx$ are, in fact, two independent terms and not part of the same formula. Thank you for making it clear. – ananta May 09 '22 at 14:57