Recently a lecturer used this notation, which I assume is a sort of twisted form of Leibniz notation:

$$y\,\mathrm{d}x - x\,\mathrm{d}y \equiv -x^2\,\mathrm{d}\left(\frac{y}{x}\right)$$

The logic here was that this could be used as:

$$\begin{align} -x^2\,\mathrm{d}\left(\frac{y}{x}\right) &\equiv -x^2\,\left(\frac{\mathrm{d}y}{x} -\frac{y}{x^2}\,\mathrm{d}x\right)\\ &\equiv y\mathrm{d}x - x\mathrm{d}y \end{align} $$

Why is this legal?

I can see some kind of differentiation going on with the second term in the above equivalence, producing the $\frac{1}{x^2}$, but having the single $\mathrm{d}$ seems like a really weird abuse of notation, and I don't quite follow why it splits the single $\frac{y}{x}$ fraction into two parts.

  • $\begingroup$ By the way, I think there's a typo: the "$y \,dy$" should be $y \,dx$. $\endgroup$ – littleO Nov 16 '20 at 22:40
  • $\begingroup$ A remark, rather than an answer. I found all this stuff confusing at UG level. But when you take a graduate geometry course and are introduced to differential forms, suddenly it makes a whole lot more sense. But that is perhaps too advanced a viewpoint for you at this juncture - start by understanding the answers as presented below, and later perhaps read about differential forms. $\endgroup$ – bounceback Nov 16 '20 at 22:59
  • $\begingroup$ @littleO thank you, that was a mistake. $\endgroup$ – jumbot Nov 16 '20 at 23:03

You should know that the differential at a point $\mathbf x_0$ of a function $\;\mathbf R^m\longrightarrow \mathbf R^n$ is the linear map $\:\ell:\mathbf R^m\longrightarrow \mathbf R^n$, that yields the best linear approximation of $f(\mathbf x_0)$ in a neighbourhood of $\mathbf x_0$, in the sense that we have $$f(\mathbf x_0+\mathbf h)=f(\mathbf x_0)+\ell(\mathbf h)+o\bigl(\|\mathbf h\|\bigr).$$ This differential is denoted $\:\mathrm d f_{\mathbf x_0}$ (or simply $\mathrm df$ for the differential at a generic point). A linear function is of course its own differential.

With the usual abuse of language that denotes a function $f$ by its value at a given point, we obtain that the differential of the $i$-th projection $\:p_i:\mathbf x=(x_1,x_2,\dots,x_m)\longmapsto x_i$ is denoted $\mathrm dx_i$.

As an example, in the case of a function of a single variable $x$, the linear map defining the differential simply corresponds to the equation of the tangent to the graph of $f$ with abscissa $x_0$: $$f(x_0+h)=f(x_0)+f'(x_0)h,\enspace\text{i.e. }\quad \mathrm df_{x_0}:h\longmapsto f'(x_0)h,$$ that we may write as $\enspace\mathrm df=f'(x)\,\mathrm dx$. This notation is generalised to functions of $m$ variables under the form $$\mathrm df=\frac{\partial f}{\partial x_1}\,\mathrm dx_1+\frac{\partial f}{\partial x_2}\,\mathrm dx_2+\dots+\frac{\partial f}{\partial x_m}\,\mathrm dx_m. $$ The usual formulæ for the derivatives have a ‘differential version’:

  • $\mathrm d(f+g)=\mathrm df+\mathrm dg$,
  • $\mathrm d(fg)=f\,\mathrm dg+g\,\mathrm df$,
  • $\mathrm d\Bigl(\dfrac fg\Bigr)=\dfrac1{g^2}\bigl(g\,\mathrm df-f\,\mathrm dg\big),$
  • $\mathrm d(g\circ f)=\mathrm dg_{f(x)}\circ\mathrm df_x$.
  • $\begingroup$ Thank you, this is a very helpful answer, in particular the last part. $\endgroup$ – jumbot Nov 16 '20 at 23:07
  • $\begingroup$ Thank you for your kind appreciation! $\endgroup$ – Bernard Nov 16 '20 at 23:08

Such arguments can always be rephrased to avoid treating $dx$ and $dy$, etc, as individual "infinitesimal" quantities. (On the other hand, "infinitesimal intuition" is a powerful and intuitive way to derive calculus formulas, so I can see why physicists are drawn to it.)

I'll assume that $y$ is a function of $x$. Let $h(x) = y(x)/x$. Then $$ h'(x) = \frac{x y'(x) - y(x)}{x^2} $$ so $$ \tag{1} -x^2 h'(x) = y(x) - x y'(x). $$ That's probably how I would write it, because the meaning is perfectly clear.

We could also write (1) using Leibniz notation: $$ -x^2 \frac{dh}{dx} = y - x \frac{dy}{dx}. $$ If we then "multiply through by $dx$", we obtain $$ - x^2 dh = y dx - x dy $$ which is what your lecturer wrote.

I can imagine that some people think the version using infinitesimal notation is more beautiful or more intuitive.

  • $\begingroup$ I am one of those people... $\endgroup$ – Ethan Bolker Nov 16 '20 at 22:28
  • $\begingroup$ @EthanBolker I see where you're coming from! $\endgroup$ – littleO Nov 16 '20 at 22:30

I believe it is being used as shorthand for: $$d\left(\frac yx\right)=dx\frac{d\left(\frac yx\right)}{dx}=dx\left(\frac{dy}{dx}x-\frac{y}{x^2}\right)=xdy-\frac{y}{x^2}dx$$

  • $\begingroup$ I think your first term in those brackets should actually be $\frac{dy}{dx}\frac{1}{x}$, but otherwise this is helpful, thank you $\endgroup$ – jumbot Nov 16 '20 at 22:16
  • $\begingroup$ Yes you are right, sorry was a typo $\endgroup$ – Henry Lee Nov 17 '20 at 1:33

In the context for this discussion you have two variables $x$ and $y$ that are somehow related. Perhaps you have $y$ as a function of $x$, or perhaps both depend on some other parameter $t$. You are interesting in knowing how small changes in $x$ and $y$ change the quotient $y/x$, written as the product $y \times (1/x)$. So what you are looking for is $d(y/x)$. If $y$ depends explicitly on $x$ you can think of this as calculating $d/dx$. If the dependence is just implicit it's easier to work with the differentials.

The actual algebra uses the product rule and the rule for differentiating $1/x$. You could do it directly with the quotient rule.

To see more intuitively what is going on, simplify the expression $$ -x^2\left( \frac{y+dy}{x+dx} - \frac{y}{x} \right) $$ and remember that $dx$ and $dy$ are small.

  • $\begingroup$ But what is it being differentiated with respect to? Surely not both $x$ and $y$? $\endgroup$ – jumbot Nov 16 '20 at 22:05
  • $\begingroup$ @JThistle See my edit. $\endgroup$ – Ethan Bolker Nov 16 '20 at 22:26
  • 1
    $\begingroup$ So in what rigorous framework are you assigning a meaning to $d(y/x)$? It isn't a differential form, so what is it? What is $d(f(x, y))$ for arbitrary $f$. I feel a great deal of sympathy for the OP if this has been presented by the lecturer as a surprise. $\endgroup$ – Rob Arthan Nov 16 '20 at 22:37
  • 1
    $\begingroup$ @RobArthan In no rigorous framework. I agree that it would be wrong to spring this kind of reasoning on beginning students. That said, I think informal reasoning with infinitesimals is too useful to be left to the physicists. $\endgroup$ – Ethan Bolker Nov 16 '20 at 22:58
  • $\begingroup$ @EthanBolker: I agree that informal reasoning methods are vital in pure mathematics as well as applied, but one needs to understand the limitations of those methods and how to translate them into rigorous mathematics when they work. $\endgroup$ – Rob Arthan Nov 16 '20 at 23:02

Very often it works to just think of $\mathrm{d}f$ as an alternative notation to $f'$ or the derivative of $f$ with respect to some unnamed variable.

For example, assume that $x$ and $y$ are functions of some common variable. Then we have $$ y\,x' - x\,y' = -x^2 (y/x)' $$ Just using the notation $\mathrm{d}f$ instead of $f'$ gives $$ y\,\mathrm{d}x - x\,\mathrm{d}y = -x^2 \mathrm{d}(y/x). $$


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