How misleading is it to regard $\frac{dy}{dx}$ as a fraction? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:57:22Z http://mathoverflow.net/feeds/question/73492 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Frank Thorne 2011-08-23T13:23:12Z 2012-10-22T03:07:40Z <p>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?</p> <p>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.</p> <p>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.</p> <p>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?</p> <p>(Note: Please don't worry, I have no intention of telling students that $\frac{dy}{dx}$ <em>is</em> a fraction; only, perhaps, that it can usually be treated as one.)</p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73496#73496 Answer by Neil Strickland for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Neil Strickland 2011-08-23T13:48:43Z 2011-08-23T18:21:56Z <p>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.</p> <p>[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 ( <a href="http://en.wikipedia.org/wiki/Maxwell_relations" rel="nofollow">http://en.wikipedia.org/wiki/Maxwell_relations</a> ) </p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73497#73497 Answer by Pietro Majer for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Pietro Majer 2011-08-23T14:08:33Z 2012-10-22T03:07:40Z <p>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). </p> <p>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. </p> <p>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 <a href="http://mathoverflow.net/questions/40082/why-do-we-teach-calculus-students-the-derivative-as-a-limit/40134#40134" rel="nofollow">here</a> on this point of view).</p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73498#73498 Answer by Steve Huntsman for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Steve Huntsman 2011-08-23T14:33:18Z 2011-08-23T14:33:18Z <p>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.</p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73501#73501 Answer by Steven Landsburg for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Steven Landsburg 2011-08-23T15:06:46Z 2011-08-23T15:06:46Z <p>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</p> <p>$$dy/dx = {\partial U/\partial x\over\partial U/\partial y}$$ by "cancelling the ${\partial U}$'s''. </p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73502#73502 Answer by John Mangual for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? John Mangual 2011-08-23T15:10:54Z 2011-08-23T15:10:54Z <p>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.</p> <p>Also real world data is not continuous time, so we are always estimating the rate of things. </p> <p>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 $\Delta y$.</p> <p>I also recommend <a href="http://www.amazon.com/Infinitesimal-Calculus-James-M-Henle/dp/0486428869/" rel="nofollow">Infinitesimal Calculus</a> by James M. Henle and Eugene M. Kleinberg as a point of view on how to teach Calc I &amp; II</p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73503#73503 Answer by Jim Conant for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Jim Conant 2011-08-23T15:11:38Z 2011-08-23T15:11:38Z <p>I am fine with using the notion of cancellation of fractions to help students <em>remember</em> 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!</p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73513#73513 Answer by David Milovich for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? David Milovich 2011-08-23T17:37:21Z 2011-08-23T17:48:07Z <p>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.</p> <p>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 <em>defined</em> 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.</p> <p>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). </p> <p>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. </p> <p>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: </p> <p>\begin{align*} \Delta y&amp;=f'(u)\Delta u+(\mathrm{small}_1)\Delta u\\ &amp;=f'(u)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)+(\mathrm{small}_1)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)\\ &amp;=f'(u)g'(x)\Delta x+(\mathrm{small})\Delta x \end{align*} No fractions here!</p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73514#73514 Answer by Mark Schwarzmann for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Mark Schwarzmann 2011-08-23T17:43:03Z 2011-08-23T17:43:03Z <p>My answer for a different question exemplifies one possible danger of taking such notation for granted: <a href="http://mathoverflow.net/questions/42929/suggestions-for-good-notation/42958#42958" rel="nofollow">Suggestions for good notation</a></p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73517#73517 Answer by Michael Hardy for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Michael Hardy 2011-08-23T17:51:35Z 2011-08-23T17:51:35Z <p>What would Euler say?</p> <p>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.</p> <p>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.</p> <p><b>Then</b> 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 <b>then</b> I point out that this is another use of the chain rule.</p> http://mathoverflow.net/questions/73492/how-misleading-is-it-to-regard-fracdydx-as-a-fraction/73542#73542 Answer by Alexander Woo for How misleading is it to regard $\frac{dy}{dx}$ as a fraction? Alexander Woo 2011-08-24T03:38:08Z 2011-08-24T03:38:08Z <p>I find $dy/dx$ misleading because it treats $x$ and $y$ as similar objects.</p> <p>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.</p> <p>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).</p> <p>Note that I am a lot more interested in (theoretical) computer science than (any form of) physics - this may bias my point of view.</p>