Consider a very simple differential equation (assuming $y$ is an invertible, continuous function and considering only real numbers, for simplicity):


If I were to solve this, I would do something like:

$$dy=dx\cdot y \implies \frac{1}{y}\cdot dy = dx$$

Then I would integrate to arrive to a solution. Of course, this works, but why? $\frac{dy}{dx}$ is not a fraction, even if it sometimes acts like one. So what are the missing steps that allow you to treat it as such? I seem to remember an explanation but I can't find it anywhere, so I'm asking here. Thanks!

  • 1
    $\begingroup$ If you use chain rule via $y = y(t)$, $x=x(t)$, then you don't have to pretend - it really is a fraction i.e. $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ the left side is not a fraction but the right side is $\endgroup$ Sep 8, 2020 at 20:39
  • $\begingroup$ Not an answer. There are several ways to justify treating that expression as a fraction. Most of them will be rather subtle. I think that at an elementary level it's probably best to treat differentials as useful in intuition and in applications - particularly true when you come to integration. (This comment is likely to generate controversy.) $\endgroup$ Sep 8, 2020 at 22:01

5 Answers 5


Everything that looks like a fraction manipulation of things is actually an entire theorem about derivatives that we are applying, often nontrivially.

  • The main thing behind doing $~\frac{dy}{dt}~\frac{dt}{dx}=\frac{dy}{dx}~,$ we are imposing the Chain Rule.
  • If we take $~\frac{1}{dy/dx}=\frac{dx}{dy}~,$ then we are using the Inverse Function Theorem.
  • Also we have to be pretty careful how we plug values into our functions.
    $(a)~$If we are integrating $~f(y(x))\frac{dy}{dx}dx~$ over some interval, and in the next step we write it as a integral of $~f(y)dy~$ over some other interval $($i.e., $u$-substitution$)$, then we are using the Fundamental Theorem of Calculus.
    $(b)~$Even Separation of Variables, from differential equations, is not given by manipulating differentials as fractions, but rather through clever use of the Chain Rule.

At no point should we pretend that $~dy/dx~$ is a fraction. What we can do is use our intuition about fractions to guide what we do when working with derivatives. Since a fraction does $~\frac{1}{a/b}=\frac{b}{a}~,$ it may be seems that "Maybe there is an analogous result about derivatives", and then find a theorem that says that $~\frac{1}{dy/dx}=\frac{dx}{dy}~,$ and see how it is precisely used (because we have to be careful how we plug stuff in). Every time we do a fraction-like manipulation of derivatives, we should think "This is because of Such-n-Such Theorem", if we don't know what theorem we are using to do the manipulation, then we shouldn't do it. At least until we find the theorem the says we can do it.

The reason why $~dy/dx~$ behave "Fraction-like", is because they are limits of things that are fractions. This does not imply that $~dy/dx~$ is a fraction, but rather that sometimes the manipulations that we do to fractions survive through the limiting process, giving us one of these main theorems. But each time, it is a nontrivial result and it is good to have all of them in the back of our mind, and to ponder about them, or whatever.

One main problem area in viewing these as fractions is when we go into higher dimensions. Here, the fraction analog breaks down almost completely. What we find is that derivatives behave more like operations from linear algebra that, in one dimension, can seem to be fraction-like, but are not so in general.

You may also see the answers from the following links:

$1.~~$ When can we use $dy/dx$ as a fraction ?
$2.~~$ How misleading is it to regard $dy/dx$ as a fraction ?
$3.~~$ Why Do People Treat $dy/dx$ as a Fraction ?
$4.~~$ Can I treat the notation $dy/dx$ as a fraction ?
$5.~~$ When not to treat dy/dx as a fraction in single-variable calculus?
$6.~~$ Why do people say dy/dx is not a fraction, but then use it as one when doing the chain rule?
$7.~~$ What is wrong with treating $\dfrac {dy}{dx}$ as a fraction?
$8.~~$ When can we not treat differentials as fractions? And when is it perfectly OK?
$9.~~$ Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?

and so on in available in the internet.

  • $\begingroup$ And what theorem are we applying if we're separating variables? $\endgroup$
    – agaminon
    Sep 9, 2020 at 10:16
  • $\begingroup$ @agaminon: Please read the full answer. Answer of your question is also there. $\endgroup$
    – nmasanta
    Sep 9, 2020 at 11:20

I think that the question has a wrong assumption. There is nothing wrong with $\frac{dy}{dx}$ actually being a fraction. Non-standard analysis gives us rigorous ways of treating infinitesimals, and therefore $\frac{dy}{dx}$ stands perfectly fine as a fraction of infinitesimals. There is nothing strange or unusual about it.

The problem most people hit is in the second derivative, but that is a failure of notation. In terms of algebraically manipulable differentials, the actual second derivative is $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. The non-functional "typical" second derivative is obtained by not treating $\frac{dy}{dx}$ as a real fraction, which is why it fails. If you do treat it as a fraction, then to get the second derivative you would use the quotient rule (since it is a quotient) and you would get the result I just stated, which is algebraically manipulable, just like the first derivative.

For more information on this, see my paper "Extending the Algebraic Manipulability of Differentials".

  • 1
    $\begingroup$ Really glad that I wasn't the only one to see this as fine from the viewpoint of non-standard analysis 😛 $\endgroup$ Mar 12, 2021 at 23:26

Accordingly, for example, Murray H. Protter, Charles B. Jr. Morrey - Intermediate Calculus-(2012) page 231 differential for function $f:\mathbb{R} \to \mathbb{R}$ is defined as function of two variables selected in special way by formula: $$df(x)(h)=f'(x)h$$ so it is linear function with respect to $h$ approximating $f$ in point $x$. Also it can be called 1-form.

This is fully rigorous definition, which does not required anything, then definition/existence of derivative. But here is more: if we define differential as existence of linear approximation in point $x=x_0$ for which holds $$f(x)-f(x_0) = A(x-x_0) + o(x-x_0), x \to x_0$$ then from this we obtain, that $f$ have derivative in point $x=x_0$ and $A=f'(x_0)$. So existence of derivative and existence of differential are equivalence requirements. Rudin W. - Principles of mathematical analysis-(1976) page 213.

If we use this definition for identity function $g(x)=x$, then we obtain $$dg(x)(h)=dx(h)=g'(x)h=h$$ This gives possibility to understand record $\frac{dy}{dx}=\frac{df}{dx}$ exactly as usual fraction of differentials and holds equality $\frac{df(x)}{dx}=f'(x)$. Exact record is $\frac{df(x)(h)}{dx(h)}=\frac{f'(x)h}{h}=f'(x)$.

When you want to integrate/solve equation $\frac{dy}{dx}=y$, then, obviously you have full rights to write $dy=ydx$ and consider both sides of equality as functions of $x$. This is same as $y'(x)=y(x)$.

Let me note, that in multivariable case this approach is not acceptable.

Addition. I cannot explain why someone assert, that $\frac{dy}{dx}$ cannot be understand as fraction - may be lack of knowledge about differential definition? For any case I bring, additionally to above source, list of books where is definition of differential which gives possibility understand fraction in question:

  1. James R. Munkres - Analysis on manifolds-(1997) 252-253 p.
  2. Vladimir A. Zorich - Mathematical Analysis I- (2016) 176 p.
  3. Loring W. Tu (auth.) - An introduction to manifolds-(2011) 34 p.
  4. Herbert Amann, Joachim Escher - Analysis II (v. 2) -(2008) 38 p.
  5. Robert Creighton Buck, Ellen F. Buck - Advanced Calculus-(1978) 343 p.
  6. Rudin W. - Principles of mathematical analysis-(1976) 213 p.
  7. Fichtenholz Gr. M - Course of Differential and Integral Calculus vol. 1 2003 240-241 p.
  8. Richard Courant - Differential and Integral Calculus, Vol. I, 2nd Edition -Interscience Publishers (1937), page 107
  9. John M.H. Olmsted - Advanced calculus-Prentice Hall (1961), page 90.
  10. David Guichard - Single and Multivariable Calculus_ Early Transcendentals (2017), page 144
  11. Stewart, James - Calculus-Cengage Learning (2016), page 190
  12. Differential in Calculus

For complete justice I mention Michael Spivak - Calculus (2008) 155 p. where author is against understanding of fractions, but argument is from kind "it is not, because it cannot be". Spivak one of my most respected and favorite authors, but "Amicus Plato, sed magis amica veritas".


Rigorously $dx$ and $dy$ can be understood only through Non-standard analysis. Then $dy$ and $dx$ are infinitesimals, a number that is smaller than any real number but is not $0$.

This is a standard formulation of first derivative in non-standard analysis. It contains infinitesimal number h, infinitesimal extension of $f(x)$ marked as $f^{*}(x)$ which behaves the same as $f(x)$ but within hyperreal numbers, which are real extended by infinitesimals. $\operatorname{st}()$ is the standard function that is converting a hyperreal into real number.


As you can see $dy$ then corresponds to $f^{*}(x+h)-f^{*}(x)$. and $dx$ to $h$, providing that the standard function $st()$ is applied implicitly before getting $f'(x)$.

  • $\begingroup$ So is there no way to understand something as simple as the separation I did without non-standard analysis? I remember someone showing a method in which you could get the same result without treating $dy$ and $dx$ as fractions. It was pretty long-winded, but the result was the same. $\endgroup$
    – agaminon
    Sep 8, 2020 at 23:28
  • $\begingroup$ As soon as you treat $dy$ and $dx$ as independent entities, you are getting into the realm of non-standard analysis. As long as you write $\frac{d}{dx}$ and keep it together, it remains not different from $f'(x)$. Another simpler explanation is that you are treating everything under the implicit $\lim_{\Delta{x} \to 0}$ and because the manipulation is always some sort of solving a linear equation and the values included are not $0$ a classical manipulation as if $dx$ and $dy$ are $\Delta{x}$ and $\Delta{y}$ works. In this case, however, there are more implicit conditions about a function. $\endgroup$
    – Alex Peter
    Sep 9, 2020 at 2:05

We can solve in that way handling $\frac{dy}{dx}$ as a fraction to obtain a solution but it is only a manipulation known as Separation of variables which needs to be justified in a rigorous way.


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