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

Question

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.)

Answer 1

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 )

Answer 2

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!

Answer 3

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''.

Answer 4

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.

Answer 5

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).

Answer 6

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 $\oint P\,dx + Q\,dy = \iint \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\, dx\, dy$ 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,

Answer 7

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.

Answer 8

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!

Answer 9

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.

Answer 10

A note from a publication (my own) that occurred several years after this question was asked. $\frac{dy}{dx}$ can be considered a fraction of differentials.

You can think of differentials as infinitesimal values that are related to each other. Non-standard analysis showed that although 19th century mathematics viewed infinitesimals as problematic, they can be easily treated as ordinary mathematical objects, capable of division, multiplication, etc.

There is no problem treating $\frac{dy}{dx}$ as a fraction, but there is a problem in higher-order derivatives and differentials, but that is because we are using a notation that doesn't support it. If you take the idea of $\frac{dy}{dx}$ being a fraction seriously, then, to find the second derivative, you are taking the derivative of a fraction. Therefore, you have to apply the quotient rule. If you apply the quotient rule to $\frac{dy}{dx}$ you do not get the typical result of $\frac{d^2y}{dx^2}$. Instead, you get: $$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$ Or, written less ambiguously: $$\frac{d(d(y))}{(d(x))^2} - \frac{d(y)}{d(x)}\frac{d(d(x))}{(d(x))^2}$$ When written this way, the second derivative can be considered actual fractions just like the first derivative. Third and higher derivatives are even uglier, because you are taking the derivative of that.

You can see more details of this in "Extending the Algebraic Manipulability of Differentials", Dynamics of Continuous, Discrete and Impulsive Systems, Series A: Mathematical Analysis 26(3):217-230, 2019. And, if anyone is concerned for its validity, it had a further review in Mathematics Magazine 92(5), pp. 396–397 in their "Reviews" section.

Answer 11

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.

Answer 12

I like the notation dy/dx because many formulas and computations become more clear and easy. However, I never thought of this as a fraction. Instead

(1) It seems to me that it would better to think of (d/dx)(y) instead of dy/dx in order to avoid misinterpretations. So, d/dx is another notation for the derivative, and df/dx is preferable to f'(x) because it points out what variable we are using. Hence, instead of the cumsy way of differentiating y=sin (x+1) by steps one can think of y=sin z, with z=x+1 and apply dy/dx= dy/dz . dz/dx. Here dy/dx only means the derivative of the function y=y(x).

(2) It then alllows us to write dy=f'(x)dx, which is coherent with using dy/dx as if it was a fraction. The formula dy=f'(x)dx is coherent with the theory of differential forms, that is, we have two 1-forms related by a function at each point. Moreover, it shows very well the idea that we have a linear function approximating the original f, with slope f'(x), but we are changing coordinates in order to put the origin at the point (x, y=y(x)).

(3) Finally, this has nothing to do with infinitesimals, excepting that the derivative dy/dx is the limit of Δy/Δx when Δx->0.

(4) Analogously for the notation udv when integrating by parts. In my opinion, the only reason for why we mathematicians prefer the notation f'(x) is because we are used to the notation $g\circ f$ for the composition of two maps, which would be another important matter for discussion.

Answer 13

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.