# Is $dy/dx$ not a ratio?

In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.

I wonder if the author say this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?

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I suppose it can within non-standard analysis. – Raskolnikov Feb 9 '11 at 16:31
In the standard formulation of calculus, dy and dx are simply not numbers, so it doesn't make sense to ask this question. – Qiaochu Yuan Feb 9 '11 at 16:38
Although it sure behaves a lot like a ratio... for instance 1/(dy/dx) = dx/dy. – user7530 Mar 3 '11 at 16:53

Historically, when Leibniz conceived of the notation, $\frac{dy}{dx}$ was supposed to be a quotient: it was the quotient of the "infinitesimal change in $y$ produced by the change in $x$" divided by the "infinitesimal change in $x$".

However, the formulation of calculus with infinitesimals in the usual setting of the real numbers leads to a lot of problems. For one thing, infinitesimals can't exist in the usual setting of real numbers! Because the real numbers satisfy an important property, called the Archimedean Property: given any positive real number $\epsilon\gt 0$, no matter how small, and given any positive real number $M\gt 0$, no matter how big, there exists a natural number $n$ such that $n\epsilon\gt M$. But an "infinitesimal" $\xi$ is supposed to be so small that no matter how many times you add it to itself, it never gets to $1$, contradicting the Archimedean Property. Other problems: Leibniz defined the tangent to the graph of $y=f(x)$ at $x=a$ by saying "Take the point $(a,f(a))$; then add an infinitesimal amount to $a$, $a+dx$, and take the point $(a+dx,f(a+dx))$, and draw the line through those two points." But if they are two different points on the graph, then it's not a tangent, and if it's just one point, then you can't define the line because you just have one point. That's just two of the problems with infinitesimals. (See below where it says "However...", though).

So Calculus was essentially rewritten from the ground up in the following 200 years to avoid these problems, and you are seeing the results of that rewriting (that's where limits came from, for instance). Because of that rewriting, the derivative is no longer a quotient, now it's a limit: $$\lim_{h\to0 }\frac{f(x+h)-f(x)}{h}.$$ And because we cannot express this limit-of-a-quotient as a-quotient-of-the-limits (both numerator and denominator go to zero), then the derivative is not a quotient.

However, Leibniz's notation is very suggestive and very useful; even though derivatives are not really quotients, in many ways they behave as if they were quotients. So we have the Chain Rule: $$\frac{dy}{dx} = \frac{dy}{du}\;\frac{du}{dx}$$ which looks very natural if you think of the derivatives as "fractions". You have the Inverse Function theorem, which tells you that $$\frac{dx}{dy} = \frac{1}{\quad\frac{dy}{dx}\quad},$$ again, almost "obvious" if you think of the derivatives as fractions. So, because the notation is so nice and so suggestive, we keep the notation even though the notation no longer represents an actual quotient, it now represents a single limit. In fact, Leibniz's notation is so good, so superior to the prime notation and to Newton's notation, that England fell behind all of Europe for centuries in mathematics and science because, due to the fight between Newton's and Leibniz's camp over who had invented Calculus and who stole it from whom (consensus is that they each discovered it independently), England's scientific establishment decided to ignore what was being done in Europe with Leibniz notation and stuck to Newton's... and got stuck in the mud in large part because of it.

(Differentials are part of this same issue: originally, $dy$ and $dx$ really did mean the same thing as those symbols do in $\frac{dy}{dx}$, but that leads to all sorts of logical problems, so they no longer mean the same thing, even though they behave as if they did).

So, even though we write $\frac{dy}{dx}$ as if it were a fraction, and many computations look like we are working with it like a fraction, it isn't really a fraction (it just plays one on television).

However... There is a way of getting around the logical difficulties with infinitesimals; this is called nonstandard analysis. It's pretty difficult to explain how one sets it up, but you can think of it as creating two classes of real numbers: the ones you are familiar with, that satisfy things like the Archimedean Property, the Supremum Property, and so on, and then you add another, separate class of reals numbers that includes infinitesimals and a bunch of other things. If you do that, then you can, if you are careful, define derivatives exactly like Leibniz, in terms of infinitesimals and actual quotients; if you do that, then all the rules of Calculus that make use of $\frac{dy}{dx}$ as if it were a fraction are justified because, in that setting, it is a fraction. Still, one has to be careful because you have to keep infinitesimals and regular real numbers separate and not let them get confused, or you can run into some serious problems.

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+1 Exceptionally good answer, very clear, very informative. – Orbling Feb 10 '11 at 3:09
As a physicist, I prefer Leibniz notation simply because it is dimensionally correct regardless of whether it is derived from the limit or from nonstandard analysis. With Newtonian notation, you cannot automatically tell what the units of $y'$ are. – rcollyer Mar 10 '11 at 16:34
@Kevin: Look at the history of math. Shortly after Newton and his students (Maclaurin, Taylor), all the developments in mathematics came from the Continent. It was the Bernoullis, Euler, who developed Calculus, not the British. It wasn't until Hamilton that they started coming back, and when they reformed math teaching in Oxford and Cambridge, they adopted the continental ideas and notation. – Arturo Magidin Mar 22 '11 at 1:42
Mathematics really did not have a firm hold in England. It was the Physics of Newton that was admired. Unlike in parts of the Continent, mathematics was not thought of as a serious calling. So the "best" people did other things. – André Nicolas Jun 20 '11 at 19:02
@BabakSorouh: Upvoting a bunch of answers of mine in a short time period, when you probably did not even have time to read them all, is not really very productive. And there is a pretty good chance that the votes will be undone by the software that checks for anomalous voting patterns in any case. Please, only up-vote that which you read and actually find worthy of the vote. – Arturo Magidin Jun 9 '12 at 4:57

Just to add some variety to the list of answers, I'm going to go against the grain here and say that you can, in an albeit silly way, interpret $dy/dx$ as a ratio of real numbers.

For every (differentiable) function $f$, we can define a function $df(x; dx)$ of two real variables $x$ and $dx$ via $$df(x; dx) = f'(x)\,dx.$$ Here, $dx$ is just a real number, and no more. (In particular, it is not a differential 1-form, nor an infinitesimal.) So, when $dx \neq 0$, we can write: $$\frac{df(x;dx)}{dx} = f'(x).$$

All of this, however, should come with a few remarks.

It is clear that these notations above do not constitute a definition of the derivative of $f$. Indeed, we needed to know what the derivative $f'$ meant before defining the function $df$. So in some sense, it's just a clever choice of notation.

But if it's just a trick of notation, why do I mention it at all? The reason is that in higher dimensions, the function $df(x;dx)$ actually becomes the focus of study, in part because it contains information about all the partial derivatives.

To be more concrete, for multivariable functions $f\colon R^n \to R$, we can define a function $df(x;dx)$ of two n-dimensional variables $x, dx \in R^n$ via $$df(x;dx) = df(x_1,\ldots,x_n; dx_1, \ldots, dx_n) = \frac{\partial f}{\partial x_1}dx_1 + \ldots + \frac{\partial f}{\partial x_n}dx_n.$$

Notice that this map $df$ is linear in the variable $dx$. That is, we can write: $$df(x;dx) = (\frac{\partial f}{\partial x_1}, \ldots, \frac{\partial f}{\partial x_n}) \begin{pmatrix} dx_1 \\ \vdots \\ dx_n \\ \end{pmatrix} = A(dx),$$ where $A$ is the $1\times n$ row matrix of partial derivatives.

In other words, the function $df(x;dx)$ can be thought of as a linear function of $dx$, whose matrix has variable coefficients (depending on $x$).

So for the $1$-dimensional case, what is really going on is a trick of dimension. That is, we have the variable $1\times1$ matrix ($f'(x)$) acting on the vector $dx \in R^1$ -- and it just so happens that vectors in $R^1$ can be identified with scalars, and so can be divided.

Finally, I should mention that, as long as we are thinking of $dx$ as a real number, mathematicians multiply and divide by $dx$ all the time -- it's just that they'll usually use another notation. The letter "$h$" is often used in this context, so we usually write $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h},$$ rather than, say, $$f'(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}.$$ My guess is that the main aversion to writing $dx$ is that it conflicts with our notation for differential $1$-forms.

EDIT: Just to be even more technical, and at the risk of being confusing to some, we really shouldn't even be regarding $dx$ as an element of $R^n$, but rather as an element of the tangent space $T_xR^n$. Again, it just so happens that we have a canonical identification between $T_xR^n$ and $R^n$ which makes all of the above okay, but I like distinction between tangent space and euclidean space because it highlights the different roles played by $x \in R^n$ and $dx \in T_xR^n$.

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Cotangent space. Also, in case of multiple variables if you fix $dx^2,\ldots,dx^n = 0$ you can still divide by $dx^1$ and get the derivative. And nothing stops you from defining differential first and then defining derivatives as its coefficients. – Alexei Averchenko Feb 11 '11 at 2:30
Well, canonically differentials are members of contangent bundle, and $dx$ is in this case its basis. – Alexei Averchenko Feb 11 '11 at 5:54
Maybe I'm misunderstanding you, but I never made any reference to differentials in my post. My point is that $df(x;dx)$ can be likened to the pushforward map $f_*$. Of course one can also make an analogy with the actual differential 1-form $df$, but that's something separate. – Jesse Madnick Feb 11 '11 at 6:18
Sure the input is a vector, that's why these linearizations are called covectors, which are members of cotangent space. I can't see why you are bringing up pushforwards when there is a better description right there. – Alexei Averchenko Feb 11 '11 at 9:12
I just don't think pushforward is the best way to view differential of a function with codomain $\mathbb{R}$ (although it is perfectly correct), it's just a too complex idea that has more natural treatment. – Alexei Averchenko Feb 12 '11 at 4:25

It is best to think of "d/dx" as an operator which takes the derivative, with respect to x, of whatever expression follows.

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It is not a ratio, just as $dx$ is not a product.

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I wonder what motivated the downvote. I do find strange that students tend to confuse Leibniz's notation with a quotient, and not $dx$ (or even $\log$!) with a product: they are both indivisible notations... My answer above just makes this point. – Mariano Suárez-Alvarez Feb 10 '11 at 0:12
I think that the reason why this confusion arises in some students may be related to the way in which this notation is used for instance when calculating integrals. Even though as you say, they are indivisible, they are separated "formally" in any calculus course in order to aid in the computation of integrals. I suppose that if the letters in $\log$ where separated in a similar way, the students would probably make the same mistake of assuming it is a product. – Adrián Barquero Feb 10 '11 at 4:09
I once heard a story of a university applicant, who was asked at interview to find $dy/dx$, didn't understand the question, no matter how the interviewer phrased it. It was only after the interview wrote it out that the student promptly informed the interviewer that the two $d$'s cancelled and he was in fact mistaken. – jClark94 Jan 30 '12 at 19:54

My favorite "counterexample" to the derivative acting like a ratio: the implicit differentiation formula for two variables. We have $$\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}$$

The formula is almost what you would expect, except for that pesky minus sign.

See http://en.wikipedia.org/wiki/Implicit_differentiation#Formula_for_two_variables for the rigorous definition of this formula.

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Typically, the $\frac{dy}{dx}$ notation is used to denote the derivative, which is defined as the limit we all know and love (see Arturo Magidin's answer). However, when working with differentials, one can interpret $\frac{dy}{dx}$ as a genuine ratio of two fixed quantities.

Draw a graph of some smooth function $f$ and its tangent line at $x=a$. Starting from the point $(a, f(a))$, move $dx$ units right along the tangent line (not along the graph of $f$). Let $dy$ be the corresponding change in $y$.

So, we moved $dx$ units right, $dy$ units up, and stayed on the tangent line. Therefore the slope of the tangent line is exactly $\frac{dy}{dx}$. However, the slope of the tangent at $x=a$ is also given by $f'(a)$, hence the equation

$$\frac{dy}{dx} = f'(a)$$

holds when $dy$ and $dx$ are interpreted as fixed, finite changes in the two variables $x$ and $y$. In this context, we are not taking a limit on the left hand side of this equation, and $\frac{dy}{dx}$ is a genuine ratio of two fixed quantities. This is why we can then write $dy = f'(a) dx$.

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The notation $dy/dx$ - in elementary calculus - is simply that: notation to denote the derivative of, in this case, $y$ w.r.t. $x$. (In this case $f'(x)$ is another notation to express essentially the same thing, i.e. $df(x)/dx$ where $f(x)$ signifies the function $f$ w.r.t. the dependent variable $x$. According to what you've written above, $f(x)$ is the function which takes values in the target space $y$).

Furthermore, by definition, $dy/dx$ at a specific point $x_0$ within the domain $x$ is the real number $L$, if it exists. Otherwise, if no such number exists, then the function $f(x)$ does not have a derivative at the point in question, (i.e. in our case $x_0$).

For further information you can read the Wikipedia article: http://en.wikipedia.org/wiki/Derivative

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So glad that wikipedia finally added an entry for the derivative...  – The Chaz 2.0 Aug 10 '11 at 17:50
I think I laugh at almost every comment you post! :) – Steve D Nov 5 '11 at 2:54
@Steve: I wish there were a way to collect all the comments that I make (spread across multiple forums, social media outlets, etc) and let you upvote them for humor. Most of my audience scoffs at my simplicity. – The Chaz 2.0 Jan 27 '12 at 21:39

$\frac{dy}{dx}$ is not a ratio - it is a symbol used to represent a limit.

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