How misleading is it to regard $\frac{dy}{dx}$ as a fraction?

I am teaching Calc I, for the first time, and I haven't seriously revisited the subject in quite some time. An interesting pedagogy question came up: How misleading is it to regard $\frac{dy}{dx}$ as a fraction?

There is one strong argument against this: We tell students that $dy$ and $dx$ mean "a really small change in $y$" and "a really small change in $x$", respectively, but these notions aren't at all rigorous, and until you start talking about nonstandard analysis or cotangent bundles, the symbols $dy$ and $dx$ don't actually mean anything.

But it gives the right intuition! For example, the Chain Rule says $\frac{dy}{du} \cdot \frac{du}{dx}$ (under appropriate conditions), and it looks like you just "cancel the $du$". You can't literally do this, but it is this intuition that one turns into a proof, and indeed if one assumes that $\frac{du}{dx} \neq 0$ this intuition gets you pretty close.

The debate about how rigorous to be when teaching calculus is old, and I want to steer clear of it. But this leaves an honest mathematical question: Is treating $\frac{dy}{dx}$ as a fraction the road to perdition, for reasons beyond the above, and which have not occurred to me?For example, what (if any) false statements and wrong formulas will it lead to?

(Note: Please don't worry, I have no intention of telling students that $\frac{dy}{dx}$ is a fraction; only, perhaps, that it can usually be treated as one.)

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Couple of related posts of possible interest @ math.se: math.stackexchange.com/questions/21199/is-dy-dx-not-a-ratio math.stackexchange.com/questions/21869/… – user11000 Aug 23 '11 at 13:41
Whether or not it can be viewed as a fraction, I think $dy/dx$ is a poor choice of notation for Calc I. The problem is illustrated by the need for the parenthetical remark "under appropriate conditions" in your third paragraph. Using $dy/dx$ means that students have to struggle with both the notion of derivative and the intricacies of the notation. Unfortunately, the notation is so prevalent that it is unreasonable to postpone the notation until Calc III or Diff Eq, where it actually comes in handy. Oh well... – François G. Dorais Aug 23 '11 at 14:21
I would argue that the notation actually leads students to make fewer mistakes. In my experience, students who do not use this notation have trouble computing the derivative of the reciprocal of a function (the formula $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$ is cumbersome compared to $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$), or performing a change of variable in an integral. – Joël Cohen Feb 29 '12 at 21:57
Teaching Calc I for the first time: Do not deviate from the textbook. Not even one tiny bit. Not even by an infinitesimal dx. – Gerald Edgar Oct 7 '13 at 20:42

You can think of $x$ and $y$ as smooth functions on a one-dimensional manifold of states of some system that you are thinking about, then $dx$ and $dy$ are differential forms. In any open region where $dx$ does not vanish we can say that $dy/dx$ is the unique smooth function such that $(dy/dx)dx=dy$; in other words, $dy/dx$ is $dy$ divided by $dx$. Of course you don't want to tell the students that, but it does clear up the logical question as asked.

[Added later:] this approach also gives a clear picture of what goes wrong with partial derivatives: if your state space has dimension $n>1$, then $dy$ and $dx$ lie in a vector space of dimension $n$, and you cannot divide them to get a number. I think it's a bit fussy to worry too much about notation for derivatives in one variable, but traditional notation for partial derivatives is horrendous, especially in any context where you might want to hold different variables constant in different places, such as Maxwell's relations in thermodynamics ( http://en.wikipedia.org/wiki/Maxwell_relations )

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There's a somewhat sneaky mistake to be made with the operator $d/dy$ (as applied to $x$). Multidimensionally, what you want to do is take directional derivatives $D_{\vec v}$. These have the property (on smooth functions) that $D_{\vec v+\vec w} = D_{\vec v} + D_{\vec w}$. If one confuses the one-form $dy$ with the vector field in the direction $y$ (by misusing the standard metric), one can get confused about whether $d/d(2y) = 1/2 d/dy$ vs. $2 d/dy$. Really, $dy$ and $d/dy$ should be sections of dual bundles, and the reciprocal suggests that correctly, as in Neil's answer. – Allen Knutson Aug 23 '11 at 14:34
Why don't you want to tell the students that? This is something that has been boggling me lately; students are already taught to think this way -- e.g. to work with dependent variables, many learn to think of $dy/dx$ as a fraction (although that is rarely the intent of the teacher), and are even taught to work with differential forms (e.g. as a device for doing integration by parts, or even change of variable). Is it really such a bad idea to actually teach this way of thinking explicitly, rather than have the students try to absorb it through osmosis? – Hurkyl Aug 24 '11 at 6:13
@Hurkyl: Well, you don't want to throw manifold theory at them at this point, which requires far more math including non-Euclidean geometry, which is serious overkill at this point. But if we are only considering functions defined on and to good ol' $\mathbb{R}$ or $\mathbb{R}^n$, is there a way to simplify the development and avoid the manifold theory? – mike3 Mar 5 at 9:34

I am fine with using the notion of cancellation of fractions to help students remember the chain rule, but it is dangerous to be too cavalier with this idea. For example, suppose $F(x,y)=0$ defines $y$ implicitly as a function of $x$. Then $$\frac{dy}{dx}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}.$$ Naive cancellation gives the wrong sign!

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This is almost an orthogonal issue; the problem is that $\frac{dy}{dx}$ along some path in XY space needn't match $\frac{\partial y}{\partial x} = 0$. – Sridhar Ramesh Aug 23 '11 at 21:27
In more detail, we have this awful notational convention for partial derivatives of not explicitly indicating which variables are being held fixed (leaving that implicit to the context), resulting in endless conflative confusion (suppose one is working with the function $f(x,y,t)=2x+3y+4t$, but looking at its values as one moves along the line $y=5x$. Should $\frac{\partial f}{\partial x}=2$ or should $\frac{\partial f}{\partial x}=2+3*5$? This ambiguity ALWAYS trips up students on multivariable chain rule problems, needlessly, due to the confusing way they're standardly written.) – Sridhar Ramesh Aug 23 '11 at 21:29
Sridhar, I suggest you write your nice comments more comprehensively in an answer! – Kevin H. Lin Aug 24 '11 at 1:57
@Sridhar: sorry, I just read this. Yes you are right that this is the fault of poor notation for the partial derivative, but the point still stands that naive cancellation is to be avoided without thinking through what it means. – Jim Conant Dec 1 '11 at 15:16
@SridharRamesh, I enjoyed your comment immensely. But can you suggest a better alternative? – goblin May 14 at 16:24

Treating dy/dx as a fraction is the gateway drug to treating ${\partial y}/{\partial x}$ as a fraction. This plus a little more notational confusion leads students to conclude that if $U(x,y)$ is a function of two variables, then along a level curve of $U$ we have

$$dy/dx = {\partial U/\partial x\over\partial U/\partial y}$$ by "cancelling the ${\partial U}$'s''.

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The problem here is the unfortunate notation for partial derivatives: the top $\partial U$ represents $dU=U_x dx+U_y dy$ with $dy$ set to zero, while the bottom $\partial U$ represents $dU$ with $dx$ set to zero; the $\partial$ represents a different differential operator in the two cases, even though it is written with the same symbol. This is somewhat orthogonal, though, to the issue of whether it is appropriate to view all these ratio-looking things as ratios; the problem isn't in treating ratio-type things as ratios, but in not notationally distinguishing different operators. – Sridhar Ramesh Aug 23 '11 at 22:02
If this argument worked, it would work equally well if you weren't moving on a level curve, but were instead following some other path. – Will Sawin Dec 24 '11 at 21:09

What would Euler say?

I tell first-year calculus students that Leibniz and Euler considered $dy$ and $dx$ to be infinitely small increments of $y$ and $x$, but that was found to be problematic in the 19th century, in more complicated problems than those considered in 1st-year calculus.

Then later I say that if $x = \tan\theta$ then $$\frac{dx}{d\theta} = \sec^2\theta = 1 + \tan^2\theta = 1 + x^2.$$ If $$\frac{dx}{d\theta} = 1 + x^2,$$ then $$\frac{d\theta}{dx} = \frac{1}{1+x^2},$$ so we have the derivative of the arctangent function.

Then I ask if anyone can say what step in the argument might be questionable. With the right very mild hints, someone will recall that $dx$ and $d\theta$ are not actual numbers, so taking reciprocals that way might be questionable. And then I point out that this is another use of the chain rule.

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If there is a well-defined tangent line to a function, then $dy/dx$ is the slope of that line, and this slope is manifestly a fraction. You can introduce (e.g.) the chain rule using this sort of thinking by noting that the slope of $y = n(mx+b)+c = mnx + (nb+c)$ is $mn$. Or the product rule by noting that the slope of $y = (mx+b)(nx+c) = mnx^2+(cm+bn)x + bc$ at $x=0$ is $mc+nb$, and by translation (but be careful here) this gives the product rule in general (as well as implying the quotient rule by replacing $nx+c$ with $-x/n+d$. No mucking around with the limit definition to get these results, just elementary analytic geometry.

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Of course you'll also have to be careful about higher-order terms to make these into proofs, but it can surely be done. – Steve Huntsman Aug 23 '11 at 14:41

I always explain in terms of linear approximation. The "derivative" of $f(x)$ is the function $f'(x)$ for which linear approximation holds, i.e. if we change $x$ to $\Delta x$ then how does $f(x)$ change? $$f(x+\Delta x) = f(x)+ f'(x)\Delta x + O(\Delta x)^2$$ The example I give my section students is $100.17^2 \approx 10034$ Do we care about the extra $0.0289$? probably not.

Also real world data is not continuous time, so we are always estimating the rate of things.

The infinitesimal point of view is useful in math an physics. One exercise is to check Green's theorem $\oint Pdx + Qdy = \int \int \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\, dx\, dy$ by integrating on/in an infinitesimal rectangle of width $\Delta x$ and height $\Delta y$.

I also recommend Infinitesimal Calculus by James M. Henle and Eugene M. Kleinberg as a point of view on how to teach Calc I & II,

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+1 for the appearance of 10034, my former Zip Code. – Gerry Myerson Aug 24 '11 at 5:41
The right remainder is $o(h)$, not $O(h^2)$ – Pietro Majer Sep 20 '15 at 7:42

A first answer to "how misleading": more than one will simplify and get $\frac{dy}{dx}=\frac{y}{x}$. A more serious objection is, thinking the derivative as the ratio of two infinitesimal increments $dy$ and $dx$ without the convenient foundation may lead a freshman student to the conclusion that every function is differentiable (if I can think to quantities $dy$ and $dx$, what's wrong in a harmless algebraic operation on them).

This does not mean one has to avoid $\frac{dy}{dx}$, but instead of using it to introduce the derivative "because it gives the right intuition", I would prefer a more rigorous definition, introducing the Leibnitz' notation only later, justifying it because it is formally consistent with the theorems about the derivatives of compositions and inverses of functions.

Personally, I prefer the definition via first order expansion: $f$ has derivative $m$ at $x$ if $f(x+h)=f(x)+mh+o(h)$ as $h\to 0$; as to the above mentioned composition rule, it is even more intuitive: the affine approximation of a composition is the composition of the affine approximations. (I happen to talk here on this point of view).

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What's most misleading about Leibnizian notation is its implicit context dependence. After you get over that hurdle, it will be easy to safely think of $dy/dx$ as a fraction.

In the context of $y=f(x)$, you think of $dx$ either as an arbitrary nonzero infinitesimal also called $\Delta x$---I did this, using Keisler's book last fall---or as a nonzero real $\Delta x$ small enough for whatever your accuracy you currently need. Either way, $dy$ is defined as $f'(x)dx$, where $f'(x)$ is defined as the usual limit of difference quotients $\Delta y/\Delta x$. Of course, in the $x=g(y)$ context, the meanings of $dx$ and $dy$ switch, as do the meanings of $\Delta x$ and $\Delta y$. In the $z=h(x,y)$ context, the meanings of $dx$, $dy$, $\Delta x$, and $\Delta y$ change yet again.

The "small enough, but not infinitely small" approach is what you'll find in standard calculus textbooks, with a section devoted to the distinction between $\Delta y$ and $dy$ (in the $y=f(x)$ context).

That said, this fall I'm planning to de-emphasize $dy/dx$ as much as I can get away with. Whether I use the little-o notation or not, I will push hard (with lots of numerical examples) on the $\Delta y=f'(x)\Delta x+o(\Delta x)$ definition of $f'(x)$, and how this makes the chain rule true but not trivial.

If $y=f(x)=x^2$ and $dx=\Delta x$ is small (but not infinitely small this time around), then $\Delta(x^2)$ equals $(x+\Delta x)^2-x^2$ equals $2x\Delta x+\Delta x^2$ equals $2x\Delta x+(\mathrm{small})\Delta x$, so $dy=2x\ dx$ and $f'(x)=2x$. In the context of $y=f(u)$ and $u=g(x)$, my presentation of the chain rule will just be that a first-order approximation of a first-order approximation is a first-order approximation:

\begin{align*} \Delta y&=f'(u)\Delta u+(\mathrm{small}_1)\Delta u\\\\ &=f'(u)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)+(\mathrm{small}_1)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)\\\\ &=f'(u)g'(x)\Delta x+(\mathrm{small})\Delta x \end{align*} No fractions here!

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I find $dy/dx$ misleading because it treats $x$ and $y$ as similar objects.

When you use this notation, you lose the important point that $y$ is a function of $x$; instead you end up looking at $x$ and $y$ as related quantities.

I think it is important for calculus students to get the idea that differentiation is an operation that takes one function and produces a new function. In that way, it is fundamentally different from addition (or unary negation) of numbers (which is not the same thing as addition of functions).

Note that I am a lot more interested in (theoretical) computer science than (any form of) physics - this may bias my point of view.

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I take it implicit differentiation is not highly regarded in Computer Science? Gerhard "I Would Have Thought Otherwise?" Paseman, 2011.08.23 – Gerhard Paseman Aug 24 '11 at 4:29
But people often really do think of $x$ and $y$ as "related quantities". And in the appropriate mode of thought, $d$ really is fundamentally the same sort of thing as addition. (but it acts to turn "differentiable numbers" into "differential forms") e.g. one way to make this precise is in the language of the category of sheaves on $\mathbb{R}$, in which "real number" really means means "continuous function $\mathbb{R} \to \mathbb{R}$. – Hurkyl Aug 24 '11 at 6:27
Gerhard - there is a right way to do implicit differentiation from this viewpoint, but it takes more space than a comment allows. – Alexander Woo Aug 24 '11 at 7:05
The "operator" notation $D_x[f]$ makes the "function of a function" idea abundantly clear. – mike3 May 21 '14 at 9:42
I do not think you should remind that y is a function of x if it is not necessary; so for me this is a feature!bug. – Anton Petrunin Jun 9 '15 at 22:34

I just wish to share, that when I was an undergrad student I felt pretty much satisfied reading the introductory chapters of the ODE book by Arnol'd.

He tackled the obscurity of $\frac{dy}{dx}$ just by defining 1-forms (in a quite understandable fashion) as linear functions on the tangent space and, furthermore, making the connection to the definition of a derivative. For me, personally, it was pretty enjoyable, as I was also a bit dazed by the notion of fractions of infinitesimally small quantities.

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My answer for a different question exemplifies one possible danger of taking such notation for granted: Suggestions for good notation

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It's VERY misleading to regard dy/dx as a mere fraction, and I believe this is one of the major pitfalls of Leibniz notation.

What it is: dy/dx is a fraction with a condition built in!

The condition is that dy is the change in y ( which we call dy ) CAUSED by a change in x ( dx ). The dy is dependent on the dx. A better way to think of dy/dx is to think of it as a function, instead, where you would plug in a dx, get an intermediate dy, and then return the ratio of dy/dx.

An example where the fraction analogy breaks down:

a) Picture two basis vectors, e1 and e2, that have the same magnitude ( but not necessarily unit-length ), and that are NOT orthogonal ( they are oblique to one another ). For the sake of visualization, let's imagine that the angle between them, theta, is equal to 60 degrees.

b) Because the vectors have the same magnitude, we can say that cos( theta ) = cos( e1, e2 ) = dy/dx = dx/dy = 0.5 ( cos(60) )

...That's right! In this situation ( |e1| = |e2| ): dy/dx = dx/dy, without necessarily having to be equal to 1. If dy/dx was a mere fraction, that would be shocking.

Why it's true: the dy in the numerator of dy/dx is not the same dy as the one in the denominator of dx/dy. The former was dependent on dx, but not the latter.

c) As an aside, for the general case, when |e1| != |e2|,

cos( theta ) = dSy / dSx = ( |e2| / |e1| ) * ( dy / dx ) = ( |e1| / |e2| ) * ( dx / dy ) = dSx / dSy

where dSx is the element of arclength along x, and equals dx * |e1| = dSx. In other words, dSx is how long in metric space, one dx in what I call "component space" amounts to ( when multiplied by |e1| ).

while dSy in dSy / dSx is how much of a change in Sy ( the metric or "length" scalar field subtanted by the y axis ) we get when we make a small change in Sx ( dSx ) along the x axis.

Once again, we run the risk of getting confused with Leibniz notation.

While we can calulate dSx = dx * |e1| and dSy = dy * |e2|, seperately, their ratio ( dSy ) / ( dSx ) is not the same as dSy/dSx = d/dSx(Sy). In dSy/dSx, the dSy is dependent on dSx, and therefore on dx.

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I don't quite understand either your explanation or your example. In fact, just as in the fraction world, $dx/dy=1/(dy/dx)$, which is usually referred to as the inverse function theorem. Overall, this looks like Marx' advice to Cauchy how to improve calculus. – Alex Degtyarev Jan 13 '15 at 10:25
That was Marx' advice :) – Alex Degtyarev Jan 13 '15 at 10:44
I never read either, though I recognize Cauchy as a famous Mathematician, so I'll take that as a compliment of sorts; even though my explanation is not coming across. What is dx? A change in the x component. Such a change in the x component ( dx ) produces a length in metric space of ( dx * |e1| = dSx ). dSy/dSx is the directional derivative along the x axis, in metric space. This ratio is equal to the cosine of the angle between the two oblique non-unit basis vectors in my example, and is same as dSx/dSy, which if the angle was 60deg = 0.5. So dSy/dSx = dSx/dSy. I can send a sketch... – Namo Jan 13 '15 at 11:00

protected by LuciaSep 19 '15 at 20:38

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