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This question regards the manipulation of derivatives as if they were fractions. But more generally it also regards doing calculus in a "Leibnizian" way.

Before asking I checked out the current poll of question regarding this topic: link, link and another one for good measure; my objective with this question is to fill in the holes that I think are left open in the discussion of this topic on this site.

In the cited questions is clarified why is incorrect to define the derivate not as a limit but as a fraction, and the consequent importance of knowing the proper definition of the derivative as a limit. This is clear. But when working whit derivatives and integrals I think the usefulness of working with Leibniz notation, and interpreting derivatives as fractions, is beyond any doubt, as this answer points out.

Problem is: when working with imprecise assumptions things can go wrong really fast, and for this reason interpreting the derivative as a fraction has the nomea of a pretty dangerous gamble.

Here are some examples on how this can go terribly wrong:
Example 1: $$\int \nabla \phi \cdot d\vec{x}=\int \frac{d\phi}{dx}dx+\frac{d\phi}{dy}dy+\frac{d\phi}{dz}dz=\int \frac{d\phi}{dx}dx+\int\frac{d\phi}{dy}dy+\int\frac{d\phi}{dz}dz=\phi+\phi+\phi=3\phi$$ but on the other hand if we state that: $d\phi+d\phi+d\phi=d\phi$ we get the correct result: $$\int \nabla \phi \cdot d\vec{x}=\int \frac{d\phi}{dx}dx+\frac{d\phi}{dy}dy+\frac{d\phi}{dz}dz=\int d\phi+d\phi+d\phi=\int d\phi=\phi$$ But if we write it with the proper notation of partial derivative then we should elide $\partial x$ with $dx$, even more chaos.
Example 2: $$\frac{\partial \phi}{\partial x}/\frac{\partial \phi}{\partial y}=\frac{\partial y}{\partial x}$$ by "eliding the $\partial \phi$".

And I am sure I could go on with more example, but I think there is no need.

Question is: is there a ruleset to follow to ensure to not fall for mistakes like this when working with Leibniz notation? I feel like this question hasn't been properly answered in the previous discussion on this topic. And if there is such ruleset, why it works?

Also it's often said that this way of handling derivatives and integrals no longer works when dealing with multivariable calculus, but from my experience it seems to give the correct answer even in this case, what is up with this exactly? Why should it no longer work? For example let's take the integral: $$\int \frac{\partial \vec{A}}{\partial t} \cdot d\vec{x}$$ It would seem a bit difficult to solve this integral "rigorously", but by interpreting it in a somewhat Leibnizian way we get: $$\int \frac{\partial \vec{A}}{\partial t} \cdot d\vec{x}=\int \frac{\partial A_x}{\partial t}dx+\frac{\partial A_y}{\partial t}dy+\frac{\partial A_z}{\partial t}dz=\int dA_x \frac{dx}{dt}+dA_y\frac{dy}{dt}+dA_z\frac{dz}{dt}=\int \vec{v} \cdot d\vec{A}=\vec{v}\cdot \vec{A}$$ This seems like a miracle, should we not use this way of working with things like this?

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    $\begingroup$ Too tired at the moment to write a full answer, but you might check out two papers of mine: "Extending the Algebraic Manipulability of Differentials" and "Exploring Alternate Notations for Partial Differentials." $\endgroup$
    – johnnyb
    Nov 20, 2020 at 0:03
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    $\begingroup$ @johnnyb I was about to comment that they might be interested in the former, and was unaware of the latter; I'm pleasantly surprised to see you on MathSE. For the former: one version can be found on the arXiv at here and another is on the site for the journal "Dynamics of Continuous, Discrete and Impulsive Systems" here. For the latter, I think it can be found here. $\endgroup$
    – Mark S.
    Nov 20, 2020 at 1:01
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    $\begingroup$ A big example that shows the dangers of playing fast and loose with Leibniz notation is: on the constraint surface $f(x,y)=c$, what is $\frac{dy}{dx}$ in terms of derivatives of $f$? Naive Leibniz notation would suggest that it is $\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$ which is off by a minus sign. $\endgroup$
    – Ian
    Nov 21, 2020 at 16:23
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    $\begingroup$ Anyway, you can do what people do in thermodynamics (which they do because of a kind of lack of a privileged representation of the relationships between quantities): think of differentials as possible changes in things with various different quantities held fixed, and then divide by them only when the constraints are consistent. So for example you can derive that "triple product rule" as $df=f_x dx + f_y dy = 0$ thus $f_y dy = -f_x dx$ and then $\frac{dy}{dx}=-f_x/f_y$, correctly. $\endgroup$
    – Ian
    Nov 21, 2020 at 16:31
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    $\begingroup$ This exposes the minus sign in an intuitive way: if $f_x$ and $f_y$ have the same sign then for $f$ to stay constant, you must change $x$ and $y$ in the opposite way, so that $f_x dx$ and $f_y dy$ can have opposite signs. $\endgroup$
    – Ian
    Nov 21, 2020 at 16:33

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The thing to keep in mind is that partial differentials, as they are commonly notated, are not distinct. That is, $\partial \phi$ can refer to any number of values. This is an unfortunate development in the history of mathematics. One way (but not the only way!) to do it is to notate on the differential which other variables were free (i.e., independent) when taking the derivative.

In this way, if $z = f(x, y)$, then $dz = \partial_x z + \partial_y z$. If I want to know the partial derivative of $z$ with respect to $x$, that is $\frac{\partial_x z}{dx}$. In your example above, $d\phi = \partial_x\phi + \partial_y\phi + \partial_z\phi$. That's why the integral is $\phi$. This is also why you can't just multiply partials and get rid of one of them. But it is why Ian's example does work.

Let's say you have $f(x, y) = c$. Then, $df = \partial_x f + \partial_yf$. The two partial derivatives, then, are $\frac{\partial_xf}{dx}$ and $\frac{\partial_yf}{dy}$. If $f(x, y) = c$, then $df = 0$, which means that $\partial_x f + \partial_yf = 0 $ which means that $\partial_yf = -\partial_x f$. Then we can rewrite the second partial derivative as $\frac{-\partial_x f}{dy}$. Then, if we put them in ratio with each other, we get: $$ \frac{\frac{\partial_xf}{dx}}{\frac{-\partial_x f}{dy}} = \frac{\partial_xf}{dx}\frac{dy}{-\partial_xf} = -\frac{dy}{dx}$$

If you rigorously keep track of the differentials, it all works out. Unfortunately, most mathematical work does not keep track so rigorously. As I mentioned in my comment, in order for all this to work out as well, you need to deal with higher-order differentials properly as well. My article "Extending the Algebraic Manipulability of Differentials" covers this. So, when taking the second derivative, you are actually taking the derivative of $\frac{dy}{dx}$. If you take differentials seriously, then this is a fraction, and therefore the quotient rule has to be applied. When you do this, you get the following formula for the second derivative: $$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$ This is not what most people use for the second derivative, but you can algebraically manipulate the differentials in this equation. Note that if you use this formula, then the "chain rule for the second derivative" becomes algebraically self-evident.

So, the question was "how do I know when I'm doing it right". I'm sorry I don't have a definitive answer, but, in general, think about what differentials are, what they mean, and where they are coming from, and only trust formulas if you know how they were derived, since many of them have historically been derived being careless with differentials.

My own hunch is that, since, in the 1800s, the validity of differential thinking was in doubt, and they all thought that derivatives were the only "actual" things, they stopped being rigorous with lone differentials.

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    $\begingroup$ What on earth is the definition of the notation $\partial_x z$ in the expression $dz=\partial_x z+\partial_y z$? The differential $dz$ is given by $\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy$. And the notation $\frac{\partial_x z}{dx}$ has no meaning. You seem to consistently omit differentials from your expressions. And finally, what is the nonsensical expression $\frac{d^2y}{dx^2}=\frac{dy}{dx}\frac{d^2x}{dx^2}$ supposed to mean?? $\endgroup$
    – Mark Viola
    Nov 22, 2020 at 18:02
  • $\begingroup$ @MarkViola - a total differential is merely the sum of its partials. As noted above, $\partial z$ does not uniquely represent a differential quantity. In the given formula, $\partial z$ could refer to one of two different differentials. The notation given is merely a way of distinguishing them. One differential if $x$ is allowed to vary ($\partial_xz$) and another if $y$ is allowed to vary ($\partial_yz$). This makes the differentials algebraically manipulable. $\endgroup$
    – johnnyb
    Nov 22, 2020 at 20:22
  • $\begingroup$ If you recognize that in the quantity $\frac{\partial f}{\partial x}$ you really mean $\frac{\partial_xf}{dx}$, then it is clear that my equation is equivalent to yours. Unclear notation is what is causing the confusion. My point is that if you use clearer notation, then the operations become clearer, and less of a "black box" operation. $\endgroup$
    – johnnyb
    Nov 22, 2020 at 20:23
  • $\begingroup$ As to $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$ (that's a subtraction, not an equal sign), this is the second derivative if you make use of algebraically manipulable differentials. Again, if you use clearer notation, the operations become clearer. For instance, if you use the standard notation of $\frac{d^2y}{dx^2}$ for the second derivative, that would naively imply that the second derivative of $y$ with respect to $t(x)$ would be $\frac{d^2y}{dx^2}\cdot \left(\frac{dx}{dt}\right)^2$. But this is clearly wrong, which you can easily test with $y = x^3$ and $x = t^2$. $\endgroup$
    – johnnyb
    Nov 22, 2020 at 20:25
  • $\begingroup$ Using the algebraically manipulable version of the second derivative gives the correct answer using basic algebraic manipulations and no special rules. $\endgroup$
    – johnnyb
    Nov 22, 2020 at 20:27

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