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 $y$?

  • $\begingroup$ 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 $\frac{d}{dx}$. $\endgroup$
    – gt6989b
    Commented May 5, 2022 at 3:45
  • $\begingroup$ 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. $\endgroup$
    – John Douma
    Commented May 5, 2022 at 3:48
  • $\begingroup$ And this is one more reason to avoid the differential notation for derivatives... $\endgroup$
    – azif00
    Commented May 5, 2022 at 3:49

2 Answers 2


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.


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.

  • $\begingroup$ 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 $\frac {dy}{dx}$ $\endgroup$ Commented May 5, 2022 at 22:35
  • $\begingroup$ Which means your answer as a whole, is wrong. $\endgroup$ Commented May 5, 2022 at 22:36
  • 3
    $\begingroup$ This 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. $\endgroup$
    – Lorago
    Commented May 5, 2022 at 22:38
  • 1
    $\begingroup$ @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 $\frac{dy}{dx}$. $\endgroup$ Commented May 6, 2022 at 11:04
  • 1
    $\begingroup$ @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. $\endgroup$ Commented May 6, 2022 at 21:11

Not the answer you're looking for? Browse other questions tagged .