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As I understand it a differential is an outdated concept from the time of Liebniz which was used to define derivatives and integrals before limits came along. As such $dy$ or $dx$ don't really have any meaning on their own. I have seen in multiple places that the idea of thinking of a derivative as a ratio of two infinitesimal change while intuitive is wrong. I understand this, and besides I am not even really sure if there is a rigorous way of saying when a quantity is infinitesimal.

Now on the other hand, it have read that you can define these differentials as actual quantities that are approximations in the change of a function. For example for a function of one real variable the differential is the function $df$ of two independent real variables $x$ and $Δx$ given by:

$df(x,Δx)=f'(x)Δx$

How this then reduces to

$df = f'(x)dx$

and again what $dx$ means I dont understand. It seems to me that it is simply a linear approximation for the function at a point $x$. However there's no mention of how large or small $dx$ must be, it seems to be just as ill defined as before and I have still found other places referring to it as an infinitesimal even when it has been redefined as here.

Anyway ignoring this, I can see how this could then be extended to functions of more than one independent variable

$y = f(x_1,....,x_n)$

$dy = $$\frac{df}{dx_1}dx_1\ +\ .... \ +\frac{df}{dx_n}dx_n\ $

However then the notion of exact and inexact differentials are brought up. This seems like its unrelated but that raise the question of what a differential means in this case.

All this comes from a course I am taking in Thermal Physics.These are the two slides[![][1]]2

If anyone can enlighten me as to what the concept of differentials means or perhaps direct me towards a book or website where I can study it myself I would be very grateful.

An explanation of Schwarz' Theorem in this context would be great too.

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  • $\begingroup$ Many words are "overloaded" in mathematical contexts. It might be helpful to imagine that the word differential is an adjective and in the case "exact differential" you ask about the noun it modifies is omitted. A phrase exact differential form would then be more easily defined/justified, but the shortened usage has been "honored" in common practice. $\endgroup$ – hardmath Jan 17 '17 at 18:25
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    $\begingroup$ I wish you lots of luck in answering your questions about differential, I've been having a hard time with them too... You seem to have already understood a lot. It's true that this notion comes from an early age of calculus and it's extremely difficult to be rigorously defined. $df(x, \Delta x) = f'(x) \Delta x$ can be written $df = f'(x) \Delta x$ by suppresing the argument of $df(x, \Delta x)$, so, if $f(x)=x$ you get $dx = \Delta x$. So, $dx = \Delta x$. In practice, this ‘definition’ of $df$ and $dx$ is a trick to enable you to speak about the derivative as a quotient. $\endgroup$ – Pythagoricus Jan 17 '17 at 18:37
  • $\begingroup$ Also, have a look at Wikipedia (‘Differential of a function’). $\endgroup$ – Pythagoricus Jan 17 '17 at 18:39
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  • Motivation

If we write $y = f(x)$, and $f'(x)$ exists for some particular $x$, then it is standard to write $\boxed{dy = f'(x) \ dx}$. If $x$ is a fixed value, then $dy$ is a dependent variable that depends on $dx$ as the independent variable.

  • Geometric Intuition

Essentially, we're introducing $dy$ and $dx$ as new coordinate axes that meet (and form their origin) at a point that lies on a curve $f(x)$. Consequently, $dy$ and $dx$ act as scales for measurement of the variables $dy$ and $dx$, just as the $x$ and $y$ axes act as scales for measurement of $x$ and $y$. In this coordinate system, you can draw the line $dy = f'(x) \ dx$ -- it is tangent to $y = f(x)$ (at some fixed $x$) and has slope $f'(x)$. The thing is that we're not defining it within the $x$-$y$ coordinate system explicitly, but rather the $dx$-$dy$ coordinate system that we constructed.

By looking at it this way, it simplifies this notion down to nothing more than recognizing the slope of a line, because from this, you may derive the popular quotient relation, $\displaystyle{\frac{dy}{dx} = f'(x)}$ for $dx \neq 0$.

  • Brief Historical Context

The $d$-notation we use for the differentials $dx$ and $dy$ goes back to Leibniz's work in the seventeenth century, but Leibniz never defined the derivative by the limit of a quotient.

  • More In-Depth Discussion

We define the differential of a function $f(x)$ as a function of $x$ as well as another independent variable $dx$ whose value is given by $df = f'(x) \ dx$. This differential is, indeed, defined at each point where $f'(x)$ exists. It's worth noting that $dx$ can take any value. For a fixed $x$, the value of the differential is just a multiple of $dx$ (since, for a fixed $x$, $f'(x)$ is a constant).

Let's re-state differentiability in the following way: a function $f$ is called differentiable at $x$ if it is defined in a neighborhood of $x$ (as well as at $x$ itself) and if there exists some $\alpha$ so that

$$\lim_{\Delta x \to 0} \frac{|f(x + \Delta x) - f(x) - \alpha \Delta x|}{|\Delta x|} = 0$$

This can be reduced to say that

$$\lim_{\Delta x \to 0} \left|\frac{f(x + \Delta x) - f(x)}{\Delta x} - \alpha \right| = 0$$

Which is to say

$$\lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = \alpha$$

Which just says that $f'(x)$ exists and is finite with value $\alpha$. Let the differential of $f$ be denoted by $df(x, dx)$. It's important to note that the dependence of $df$ on $x$ is not the same as its dependence on $dx$, so perhaps the notation I used is not quite clear. We're noting that $df$ depends linearly on $dx$. We can do some re-writing:

$$\lim_{\Delta x \to 0} \frac{|f(x + \Delta x) - f(x) - \alpha \Delta x|}{|\Delta x|} = \lim_{dx \to 0} \frac{|f(x + dx) - f(x) - df(x, dx)|}{|dx|}$$

because $\alpha \ dx = f'(x) \ dx = df(x, dx)$ (note that, for notational purposes, I've replaced $\Delta x$ with $dx$, since $dx$ can take any value). So, we've learned that, for a fixed $x$, $df$ is a multiple of $dx$ and that the limit relation we established above is valid. What the relation above tells us is that $df(x, dx)$ is a decent approximation to $f(x + dx) - f(x)$ in the sense that the difference $f(x + dx) - f(x)-df(x,dx)$ is very small compared to $|dx|$ as $|dx| \to 0$.

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  • $\begingroup$ Some post scripts: It's worth noting that $dx$ and $dy$ don't necessarily have to be small for the relation $dy/dx = f'(x)$ to be valid. An exact differential would perhaps require another whole discussion. But simply, something like $dA = B(x, y, z) \ dx + C(x, y, z) \ dy + D(x, y, z) \ dz $ is an exact differential (form) if $\left<B, C, D \right>$ is a conservative vector field with potential function $A$. From this, it follows that $\int dA$ is not path-dependent. $\endgroup$ – Alex Menendez Jan 17 '17 at 19:45
  • $\begingroup$ From a bit of research it seems that you are talking about the concept of differential forms. Am I correct in saying that? $\endgroup$ – RonGiant Jan 18 '17 at 10:20
  • $\begingroup$ $B(x, y, z) \ dx + C(x, y, z) \ dy + D(x, y, z) \ dz$ is a differential form, and is said to be exact if there exists a scalar function $A$ so that $dA = B(x, y, z) \ dx + C(x, y, z) \ dy + D(x, y, z) \ dz$ -- the differentials are $dA$, $dx$, $dy$, and $dz$. So, when we say that $B(x, y, z) \ dx + C(x, y, z) \ dy + D(x, y, z) \ dz$ is an exact differential form, it's equivalent to saying that the differential $dA$ is exact (granted such an $A$ exists) -- that's where we develop the notion of the "exact differential" from. $\endgroup$ – Alex Menendez Jan 18 '17 at 18:06
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You can think of differential (1-forms) as new, independent variables. If $y=f(x)$, then it's convenient to place the origin of the $dx-dy$ axes on the curve given by $y=f(x)$. The new variables are related to $y$ and $x$ by the equation $dy = f'(x)\; dx$. (This is the equation of the tangent line at that point, but given in $dx-dy$ coordinates.) By leaving $dx$ and $dy$ as variable, it's possible to let $dx$ take on any value you like, e.g., $\Delta x$. Then the linear approximation theorem says that if $dx = \Delta x$, then $\Delta y \approx dy.$ And we can see that from the picture described above. This can be extended to functions of more than one variable.

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One dimensional calculus is very special. For example

  • If $f$ is any continuous function, then $f(x) \, \mathrm{d}x$ has an antiderivative
  • If $y$ and $u$ depend differentiably on $x$, then $\mathrm{d}y$ and $\mathrm{d}u$ are proportional to each other. This means it makes sense to speak of the ratio $\frac{\mathrm{d}y}{\mathrm{d}u}$ (at least wherever $\mathrm{d}u$ is 'nonzero')
  • To talk about definite integrals $\int_a^b f(x) \, \mathrm{d}x$ requires that we only need to specify the endpoints

In higher dimensions, things don't work out so nicely; when we do have these nice properties, it is a special thing deserving a name.

Exactness is the property of being antidifferentiable:

A differential $\omega$ is exact if and only if there exists a function $f$ such that $\omega = \mathrm{d}f$

It turns out that exact forms satisfy the fundamental theorem of calculus: if $\gamma$ is a path from $P$ to $Q$, then

$$ \int_\gamma \mathrm{d}f = f(Q) - f(P) $$

So hopefully you can see exactness being very important.


We can define a product on differentials called the wedge product; it sort of behaves like you'd expect a product to behave, except that it has the weird feature that it's antisymmetric on ordinary differentials: $\mathrm{d}x \wedge \mathrm{d}y = -\mathrm{d}y \wedge \mathrm{d}x$. (and also $\mathrm{d}x \wedge \mathrm{d}x = 0$).

W can use this to differentiate a differential form: to get a 2-form:

$$ \mathrm{d}\left( u \, \mathrm{d}v \right) = \mathrm{d}u \wedge \mathrm{d} v $$ and furthermore to even higher degrees.

For example, in $\mathbb{R}^2$, we could compute $$\begin{align} \mathrm{d}\left( f(x,y) \, \mathrm{d}x + g(x,y) \, \mathrm{d} y \right) &= \mathrm{d}f(x,y) \wedge \mathrm{d}x + \mathrm{d}g(x,y) \wedge \mathrm{d} y \\&= (f_1(x,y) \mathrm{d}x + f_2(x,y) \mathrm{d}y) \wedge \mathrm{d}x + (g_1(x,y) \mathrm{d}x + g_2(x,y) \mathrm{d}y) \wedge \mathrm{d}y \\&= (g_1(x,y) - f_2(x,y)) \mathrm{d}x \wedge \mathrm{d}y \end{align}$$

(I use $h_n$ to refer to the partial derivative of a function with respect to its $n$-th argument while holding the other arguments constant)

You might notice that the test using Schwarz' rule that your image is alluding to boils down to checking if a differential $\omega$ satisfies $\mathrm{d} \omega = 0$. These are also special:

A differential form $\omega$ is closed if and only if $\mathrm{d} \omega = 0$

It turns out that every exact differential form is closed, and conversely, in extremely nice spaces like $\mathbb{R}^n$, every closed differential form is exact.

Since testing if a form is closed is relatively easy and computational, it provides a very nice first step in checking whether a differential is exact — and in very nice spaces like $\mathbb{R}^n$ it's the only step you need!

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