I have been forewarned about it, I have read the answers here, but I haven't seen a counter example where it doesn't work. I know that it isnt really a fraction, but does it effectively get the same result in all cases, or are there counterexamples I must be warned about.

I do not mean obvious counter examples, like $$\frac{dy}{dx}+\frac{du}{dv} = \frac{dydv+dxdu}{dx dv}$$ which as far as I know doesn't really mean anything.

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    $\begingroup$ I would have thought that having something go "obviously wrong" would be better than something going "wrong incognito"... $\endgroup$ – Arturo Magidin Jun 20 '11 at 18:50
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    $\begingroup$ According to non-standard calculus (en.wikipedia.org/wiki/Non-standard_calculus) you can think of a derivative as the 'standard part' of a ratio of infinitesimals. $\endgroup$ – George Lowther Jun 20 '11 at 18:56
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    $\begingroup$ Note: $d(x^2)/dx=((x+dx)^2-x^2)/dx=2x+dx\not=2x$. You have to neglect the additional $dx$ term to get the derivative, which is why you have to look at the standard part of the ratio, rather than just the ratio itself. $\endgroup$ – George Lowther Jun 20 '11 at 18:59
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    $\begingroup$ Would implicit function theorem count? It gives a "wrong" sign :) $\endgroup$ – Marek Jun 20 '11 at 19:01
  • $\begingroup$ What about $\frac{d^2 y}{dx^2}$? $\endgroup$ – André Nicolas Jun 20 '11 at 19:06

$\frac{dy}{dx}$ can indeed be thought of as a ratio. The question is, what do $dx$ and $dy$ represent in this ratio? The answer is, changes in $x$ and $y$ along the tangent line to the curve at the point in question, rather than along the curve itself. See e.g. https://www.encyclopediaofmath.org/index.php/Differential


The chain rule in two dimensions is counter-intuitive at first if you think of the derivative (and partial derivatives) as ratios:

$$ \frac{\mathrm d f}{\mathrm dt} = \frac{\partial f}{\partial x} \frac{\mathrm dx}{\mathrm dt} + \frac{\partial f}{\partial y} \frac{\mathrm dy}{\mathrm dt}. $$

Here, $f$ is a function of two variables, $x$ and $y$, both of which are functions of $t$.

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    $\begingroup$ This is because the partial derivative notation is bad. From an infinitesimals point of view, $df = \partial_x f + \partial_y f$, where $\partial_x f$ is the numerator of the first partial derivative, and $\partial_y f$ is the numerator of the second partial derivative. $\endgroup$ – Jim Belk Jun 21 '11 at 1:42
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    $\begingroup$ Not at all counter-intuitive. Here's a picture to illustrate $dz = \frac{\partial z}{\partial x} dx + \frac{\partial f}{\partial y} dy$: link $\endgroup$ – Robert Israel Jun 21 '11 at 1:48
  • $\begingroup$ For posterity, @RobertIsrael's link is mirrored at i.stack.imgur.com/JtCYx.jpg $\endgroup$ – wchargin Dec 25 '17 at 8:01

This question has come up at MathOverflow:


and overlaps with other questions on this site:

Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?

If $\frac{dy}{dt}dt$ doesn't cancel, then what do you call it?

There are several interesting answers at each of those links.


The rule of thumb to use is to only think in this manner when there is only one independent variable (e.g. when all of your variables are to be thought of as functions of $x$). When all of the variables you are using depend on each other, there a variety of (fully rigorous) ways to think of expressions like "$dx$" as simply being a value with funny units, which cancel out in "$\frac{dy}{dx}$", resulting in a perfectly ordinary value.

But when there are more variables (e.g. when considering $z = f(x,y)$), this way of thinking no longer works. (Or more accurately, it does still work, but $dx, dy, dz$ behave like vectors so division doesn't make sense)


protected by J. M. is a poor mathematician Nov 5 '11 at 3:49

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