I ask because, as a first-year calculus student, I am running into the fact that I didn't quite get this down when understanding the derivative:

So, a derivative is the rate of change of a function with respect to changes in its variable, this much I get.

Thing is, definitions of 'differential' tend to be in the form of defining the derivative and calling the differential 'an infinitesimally small change in x', which is fine as far it goes, but then why bother even defining it formally outside of needing it for derivatives?

And THEN, the bloody differential starts showing up as a function in integrals, where it appears to be ignored part of the time, then functioning as a variable the rest.

Why do I say 'practical'? Because when I asked for an explanation from other mathematician parties, I got one involving the graph of the function and how, given a right-angle triangle, a derivative is one of the other angles, where the differential is the line opposite the angle.

I'm sure that explanation is correct as far it goes, but it doesn't tell me what the differential DOES, or why it's useful, which are the two facts I need in order to really understand it.

Any assistance?

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    Although not directly addresssing your questions, you might want to look at this answer http://math.stackexchange.com/questions/21199/dy-dx-is-not-a-ratio/21209#21209 and this one http://math.stackexchange.com/questions/21869/if-fracdydtdt-doesnt-cancel-then-what-do-you-call-it/21876#21876 – Arturo Magidin Feb 26 '11 at 21:52
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    Anyone who sees calculus in application is likely to encounter both derivatives and differentials. The two concepts have confusingly similar notation. For that reason, this post is a very important contribution. –  Jun 04 '12 at 04:24
  • [Relevant](http://math.stackexchange.com/questions/852394/failure-of-differential-notation/1651264#1651264) –  Feb 11 '16 at 23:41
  • Practically, the difference is infinitely small. – Torsten Schoeneberg Mar 06 '22 at 06:41

6 Answers6


Originally, "differentials" and "derivatives" were intimately connected, with derivative being defined as the ratio of the differential of the function by the differential of the variable (see my previous discussion on the Leibnitz notation for the derivative). Differentials were simply "infinitesimal changes" in whatever, and the derivative of $y$ with respect to $x$ was the ratio of the infinitesimal change in $y$ relative to the infinitesimal change in $x$.

For integrals, "differentials" came in because, in Leibnitz's way of thinking about them, integrals were the sums of infinitely many infinitesimally thin rectangles that lay below the graph of the function. Each rectangle would have height $y$ and base $dx$ (the infinitesimal change in $x$), so the area of the rectangle would be $y\,dx$ (height times base), and we would add them all up as $S\; y\,dx$ to get the total area (the integral sign was originally an elongated $S$, for "summa", or sum).

Infinitesimals, however, cause all sorts of headaches and problems. A lot of the reasoning about infinitesimals was, well, let's say not entirely rigorous (or logical); some differentials were dismissed as "utterly inconsequential", while others were taken into account. For example, the product rule would be argued by saying that the change in $fg$ is given by $$(f+df)(g+dg) -fg = fdg + gdf + df\,dg,$$ and then ignoring $df\,dg$ as inconsequential, since it was made up of the product of two infinitesimals; but if infinitesimals that are really small can be ignored, why do we not ignore the infinitesimal change $dg$ in the first factor? Well, you can wave your hands a lot of huff and puff, but in the end the argument essentially broke down into nonsense, or the problem was ignored because things worked out regardless (most of the time, anyway).

Anyway, there was a need of a more solid understanding of just what derivatives and differentials actually are so that we can really reason about them; that's where limits came in. Derivatives are no longer ratios, instead they are limits. Integrals are no longer infinite sums of infinitesimally thin rectangles, now they are limits of Riemann sums (each of which is finite and there are no infinitesimals around), etc.

The notation is left over, though, because it is very useful notation and is very suggestive. In the integral case, for instance, the "dx" is no longer really a quantity or function being multiplied: it's best to think of it as the "closing parenthesis" that goes with the "opening parenthesis" of the integral (that is, you are integrating whatever is between the $\int$ and the $dx$, just like when you have $2(84+3)$, you are multiplying by $2$ whatever is between the $($ and the $)$ ). But it is very useful, because for example it helps you keep track of what changes need to be made when you do a change of variable. One can justify the change of variable without appealing at all to "differentials" (whatever they may be), but the notation just leads you through the necessary changes, so we treat them as if they were actual functions being multiplied by the integrand because they help keep us on the right track and keep us honest.

But here is an ill-kept secret: we mathematicians tend to be lazy. If we've already come up with a valid argument for situation A, we don't want to have to come up with a new valid argument for situation B if we can just explain how to get from B to A, even if solving B directly would be easier than solving A (old joke: a mathematician and an engineer are subjects of a psychology experiment; first they are shown into a room where there is an empty bucket, a trashcan, and a faucet. The trashcan is on fire. Each of them first fills the bucket with water from the faucet, then dumps it on the trashcan and extinguishes the flames. Then the engineer is shown to another room, where there is again a faucet, a trashcan on fire, and a bucket, but this time the bucket is already filled with water; the engineer takes the bucket, empties it on the trashcan and puts out the fire. The mathematican, later, comes in, sees the situation, takes the bucket, and empties it on the floor, and then says "which reduces it to a previously solved problem.")

Where were we? Ah, yes. Having to translate all those informal manipulations that work so well and treat $dx$ and $dy$ as objects in and of themselves, into formal justifications that don't treat them that way is a real pain. It can be done, but it's a real pain. Instead, we want to come up with a way of justifying all those manipulations that will be valid always. One way of doing it is by actually giving them a meaning in terms of the new notions of derivatives. And that is what is done.

Basically, we want the "differential" of $y$ to be the infinitesimal change in $y$; this change will be closely approximated to the change along the tangent to $y$; the tangent has slope $y'(a)$. But because we don't have infinitesimals, we have to say how much we've changed the argument. So we define "the differential in $y$ at $a$ when $x$ changes by $\Delta x$", $d(y,\Delta x)(a)$, as $d(y,\Delta x)(a) = y'(a)\Delta x$. This is exactly the change along the tangent, rather than along the graph of the function. If you take the limit of $d(y,\Delta x)$ over $\Delta x$ as $\Delta x\to 0$, you just get $y'$. But we tend to think of the limit of $\Delta x\to 0$ as being $dx$, so abuse of notation leads to "$dy = \frac{dy}{dx}\,dx$"; this is suggestive, but not quite true literally; instead, one then can show that arguments that treat differentials as functions tend to give the right answer under mild assumptions. Note that under this definition, you get $d(x,\Delta x) = 1\Delta x$, leading to $dx = dx$.

Also, notice an interesting reversal: originally, differentials came first, and they were used to define the derivative as a ratio. Today, derivatives come first (defined as limits), and differentials are defined in terms of the derivatives.

What is the practical difference, though? You'll probably be disappointed to hear "not much". Except one thing: when your functions represent actual quantities, rather than just formal manipulation of symbols, the derivative and the differential measure different things. The derivative measures a rate of change, while the differential measures the change itself.

So the units of measurement are different: for example, if $y$ is distance and $x$ is time, then $\frac{dy}{dx}$ is measured in distance over time, i.e., velocity. But the differential $dy$ is measured in units of distance, because it represents the change in distance (and the difference/change between two distances is still a distance, not a velocity any more).

Why is it useful to have the distinction? Because sometimes you want to know how something is changing, and sometimes you want to know how much something changed. It's all nice and good to know the rate of inflation (change in prices over time), but you might sometimes want to know how much more the loaf of bread is now (rather than the rate at which the price is changing). And because being able to manipulate derivatives as if they were quotients can be very useful when dealing with integrals, differential equations, etc, and differentials give us a way of making sure that these manipulations don't lead us astray (as they sometimes did in the days of infinitesimals).

I'm not sure if that answers your question or at least gives an indication of where the answers lie. I hope it does. Added. I see Qiaochu has pointed out that the distinction becomes much clearer once you go to higher dimensions/multivariable calculus, so the above may all be a waste. Still...

Added. As Qiaochu points out (and I mentioned in passing elsewhere), there are ways in which one can give formal definitions and meanings to infinitesimals, in which case we can define differentials as "infinitesimal changes" or "changes along infinitesimal differences"; and then use them to define derivatives as integrals just like Leibnitz did. The standard example of being able to do this is Robinson's non-standard analysis Or if one is willing to forgo looking at all kinds of functions and only at some restricted type of functions, then you can also give infinitesimals, differentials, and derivatives substance/meaning which is much closer to their original conception.

Arturo Magidin
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    It's worth mentioning that there are ways to make sense of infinitesimals, e.g. using nonstandard analysis (http://en.wikipedia.org/wiki/Non-standard_analysis) or in the restricted setting of polynomials. – Qiaochu Yuan Feb 26 '11 at 22:50
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    In the version of that joke I originally heard, the second "problem" had the faucet already running and making a mess, which the mathematician reduces to a solved problem by (of course) setting the trash on fire. – camccann Dec 28 '12 at 20:09
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    The first answer is using so many other complex mathematical terms that it does not help in any way. Unless you are a genius in advanced linear algebra ( at that point, you would not even need a clarification between a differential and a derivative. you should know it) However, the second one was the best explanation I've read by far. Monsieur Magidin, Please come teach in Canada! –  Jan 18 '13 at 04:15
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    Enlightening explanation, thank you so much! One concern: "tend to give the right answer under mild assumptions. Note that under this definition, you get.. leading to dx = dx." Should that be dx = ∆x, or am I completely missing the point? – electronpusher Dec 23 '16 at 02:30
  • when you say " Differentials were simply "infinitesimal changes" in whatever" you mean it was changes in function or variable? i.e not any specific kind of object right? also does new modern defintion of differenital differ from the old one?https://en.wikipedia.org/wiki/Differential_of_a_function – user123124 Oct 21 '18 at 13:06
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    @user1: Yes, of course “whatever” doesn’t mean “horses”, “cars”, “airplanes”. It means objects of study, which were **quantities** (they could be variables, or values of functions; keep in mind that the modern notion of “function” didn’t even exist yet; it was first introduced by Euler, and refined over decades after that). Yes, the modern notion is different, because today we don’t have infinitesimals outside of nonstandard analysis; the old definition is nonsense in the context of the modern construction of the real numbers. – Arturo Magidin Oct 21 '18 at 17:18
  • @ArturoMagidin - I have a rather basic question after reading your post regarding the wording *with respect to*. I often see the derivative described as follows: *the derivative of a function is the rate of change of the value of the function with respect to its argument*. I have never thought about it before, but when we use the wording *with respect to*, does this really mean: [the derivative of a function is] the rate of change of the value of the function *as* the argument changes? – Taylor Rendon Jan 07 '21 at 17:33
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    @TaylorRendon: It is the (limit) of the ratio of change in the function by the change of the argument. It's a *relative* change, a *rate*, not an absolute change. So, yes, it means exactly that. – Arturo Magidin Jan 07 '21 at 17:39
  • @ArturoMagidin Sorry, I know this is an old answer but you say, a way to make sense out of differentials is to define them as $d(y, \Delta x) = y' dx$ and that we'd want to think of $\lim\limits_{\Delta x \to 0} \Delta x = dx$. Isn't the limit simply $0$ and so $dx =0$? I know you mentioned you've abused some notation but something here doesn't seem... right at all. How are you defining a differential in terms of a differential? Isn't this circular reasoning? Why go through the pain of defining differentials only to end up with saying $dx = \Delta x$. What did I miss? – William Aug 23 '21 at 14:20
  • @William: that very paragraph says that literally speaking, that just leads to $dx=dx$ and to $dx=0$. Again, "differentials" don't really work in the absence of infinitesimals, and infinitesimals don't work in the context of the usual real numbers. Mostly, they work only as formal manipulations, and in fact modern treatment of calculus does not really get into differentials at all for precisely that reason. I also note that you misstate what I said for the "differential of $y$ at $a$ when $x$ changes by $\Delta x$. It's not "$y'\,dx$", it's $y'(a)\Delta x$. – Arturo Magidin Aug 23 '21 at 16:02
  • Thank you for the reply. I'm confused by your line "Again, "differentials" don't really work... modern treatment of calculus does not really get into differentials at all for precisely that reason". Unless I'm misunderstanding, are you saying there is no way to rigourously define Differentials in $\mathbb{R}$? So when there are instances in real analysis, say while substitution in integrals. What am I really doing? I'm painfully bothered by not being able to make sense out of differentials. I've read so many answers here and none of them seem convincing honestly. What do you suggest I do? :( – William Aug 23 '21 at 18:36
  • @William: There is no consistent way to define differentials that does not reduce to $dx$ being just another symbol for $\Delta x$, $dy$ being a symbol for $y'\Delta x$, and it *not* having the properties that differentials used to have. If you want to know what you are "really" doing when you do a substitution in integrals, you need to learn about the Jacobian and actual, honest, formal real analysis, not Calculus, not math.stackexchange, and not a comment thread. The best you can do is think of it as a notational trick that makes things work. Or learn some **real** real analysis. – Arturo Magidin Aug 23 '21 at 19:25
  • Thank you for replying. I'm actually familiar with linear algebra and real analysis and multivariate. I was not referring to the dumb high school calculus. I could formally define derivatives, integrals but not differentials. Where do you suggest I begin, specifically to learn about differentials. (Not too advanced, I'm only undergrad and I'm quite comfortable with most undergrad subjects.) – William Aug 23 '21 at 22:31
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    @William: Make up your mind. Either you can take an actual, honest, real, formal real analysis class because you are "actually familiar with linear algebra and real analysis", or you can't because it is "too advanced". I pointed you in the direction of the Jacobian. But if you aren't able or willing to go "too advanced", then sorry, but you are plain out of luck. If you want to learn the formal side of differentials, you're going to have to wait until you are ready and willing to take an actual advanced course. So, I don't think there's anything else to talk about here. – Arturo Magidin Aug 24 '21 at 00:47
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    @William: Just to get you an idea, you can look at the "Other approaches" section on the [Wikipedia page](https://en.wikipedia.org/wiki/Differential_of_a_function?wprov=sfti1). They can be formally defined as the exterior derivative of a function; that is done in the context of differential geometry, which takes quite a bit of material before you can get into it. Not every question that can be formulated simply has a simple answer, or even an answer that can be encompassed in this site (let alone a comment thread). – Arturo Magidin Aug 24 '21 at 00:55

The distinction between the two concepts is not really clarified until you move up to higher dimensions and start doing multivariable calculus. Derivatives become Jacobians, which are matrices giving linear approximations to smooth functions at a point, while differentials become differential forms, which are things you integrate over higher-dimensional regions. (This is why we bother with differentials.)

In one dimension it is hard to tell the difference between these two concepts because the Jacobian (derivative) $f'(x)$ looks an awful lot like the differential form $f'(x) \, dx$. But in higher dimensions the distinction becomes clearer: the relationship between the Jacobian and differential forms is given by the multivariate change of variables formula.

Edit: to be completely precise, the problem is that in one dimension there is only one space of differential forms and the action of the Jacobian on it doesn't look special at all. In $n$ dimensions there are $n$ spaces of differential forms, and the Jacobian acts differently on each of them. In particular, on the highest-dimension forms, it acts by the determinant, which is where the change of variables formula comes from.

Qiaochu Yuan
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A 1000-words' answer is not required to explain this (as other answers).

See this answer in Quora: What is the difference between derivative and differential?.

In simple words, the rate of change of function is called as a derivative and differential is the actual change of function.

We can also define a derivative in terms of differentials as the ratio of differentials of function by the differential of a variable.

A derivative is the change in a function ($\frac{dy}{dx}$); a differential is the change in a variable$ (dx)$.

A function is a relationship between two variables, so the derivative is always a ratio of differentials.

I think this is the best explanation so far.

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  • If the derivative is $\frac{\mathrm{d}y}{\mathrm{d}x}$, then the differential should be $\mathrm{d}y$ and not $\mathrm{d}x$. Or more precisely $\rightarrow$ Let $y = f(x)$, then $\frac{\mathrm{d}y}{\mathrm{d}x} = f'(x)$ is the derivative, and the differential would be $\mathrm{d}y = f'(x)\mathrm{d}x$. – x.projekt Feb 17 '22 at 12:56

Not for basic calculus. Short answer is that derivatives are result of applying an element of the tangent space or a vector space to a a real valued function. While a diferential is a result of a map between manifolds or a diferential form. In the special case where M,N are Euclidian m space and R those are mostly the same except the notation.

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These answers haven't formalized the objects $dx$, so I'll give my own answer which does. This will be more high level and requires some understanding of linear algebra, group actions, and calculus in more than one variable. There's a few objects we need to make clear: the basis vector $\frac{\partial}{\partial x}$, the projection map $x^i$, and the operator d.

The setup for our story is going to be in affine space $\mathbb{A}^n$ which is defined as $\mathbb{R}^n$ with a simply transitive group action on itself by translations. We say that $\mathbb{A}^n$ is defined over $\mathbb{R}^n$ and think of elements in the former as points and the elements in the latter as vectors. The upshot of the words "simply transitive" is that subtracting two points always gives a unique vector. If these words are unfamiliar, feel free to think of $\mathbb{A}^n$ as $\mathbb{R}^n$ + translations. We define an affine subspace $A\subset \mathbb{A}^n$ to be a subset with a simply transitive group action from a linear subspace $V\subset \mathbb{R}^n$. We call $V$ the tangent space of $A$. Note that $\mathbb{R}^n$ comes with canonically equipped with a set of basis vectors $\{e_i\}$ with $e_i=(0,...,1,...0)$ where the 1 is in the $i$th position. We rename $\frac{\partial}{\partial x^i}:=e_i$. We also have the standard projection maps $x^i:\mathbb{R}^n\to \mathbb{R}$ by remembering only the $i$th coordinate. We define $dx^i\in (\mathbb{R}^n)^* $ to be the dual to vector $\frac{\partial}{\partial x^i}$, so $\{dx^i\}$ forms a basis for $(\mathbb{R}^n)^*$. Since we built $\mathbb{A}^n$ on top of $\mathbb{R}^n$, we also get a canonical basis and projection maps on $\mathbb{A}^n$.

For example, fix $\mathbb{A}^2$ and let $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ denote the basis vector for the underlying tangent space $\mathbb{R}^2$. The graph of equation $y=3$ is an affine subspace of $\mathbb{A}^2$ with tangent space the span of $\frac{\partial}{\partial x}\in \mathbb{R}^2$. Another example would be the graph of the equation $y=x$ is an affine subspace with tangent space the span of $\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$.

At this point in our story we require that $\mathbb{R}^n$ has a norm denoted $\|\cdot\|$ which induces a distance function on $\mathbb{A}^n$ via subtraction. Let $U\subset \mathbb{A}^n$ be a open set, and $p\in U$ a point. We say $f:U\to \mathbb{A}^m$ is differentiable at $p$ if there exists a linear function which we denote $df_p:\mathbb{R}^n\to \mathbb{R}^m$ which satisfies the following inequality: For every $\epsilon>0$, there exists a $\delta$ such that if $\|\xi\|<\delta$, then $$\|f(p+\xi)-f(p)-df_p(\xi)\|<\epsilon \|\xi\|$$

We say $f$ is differentiable on $U$ if it's differentiable at every $p\in U$. One may call $df_p$ the derivative at $p$ in a multivariable class. Fix codomain $\mathbb{A}^1$. We don't lose generality by restricting the codomain to one dimensions becuase every function $f:\mathbb{A}^n\to \mathbb{A}^m$ can be expressed as coordinate functions $f=(f^1,...,f^n)$ by composing with projection map $x^i$. Let me denote Diff($U$,$\mathbb{A}^1$) to be the set of differentiable functions between the two, and Map($U$,$\mathbb{A}^1$) the set of all functions between the two. These two form a vector space structure by function addition and scalar multiplication defined in the usual way. We define an operator between the these two vector spaces

$$\mathfrak{d}:\text{Diff}(U,\mathbb{A}^1)\to \text{Map}(U,\mathbb{A}^1)$$ We require $\mathfrak{d}$ to be linear and obey the Leibniz rule, namely given $f,g\in \text{Diff}(U,\mathbb{A}^1)$ $$\mathfrak{d}(fg)=\mathfrak{d}f \cdot g+ f\cdot \mathfrak{d}f$$

We call $\mathfrak{d}$ a derivation if it satisfies the above conditions, and we see that the differential operator $d$ is a special case of this. So it turns out that taking $U=\mathbb{A}^n$ and $f=x^i$ the projection map, $dx^i$ which we defined as the dual basis to $\frac{d}{dx^i}$ is actually the differential of $x^i$ (I'll leave this to whoever the hell read this far to show). There's a whole lot more to be said and skimmed over. If you're interested, all this information and more comes from here.

Zack Fox
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The limit is the value corresponding to a function (say $f(x)$) at a specific point (say $x = x_0$). The differential of $f$ is a formula in which the limit-notion is used.

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