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This previous question of mine has lead me to ask the following question:

It was my understanding that the chain rule $$\dfrac{du}{dx}=\dfrac{dy}{dx}\dfrac{du}{dy}$$ only makes sense when there is some function $u$ for it to operate on.

So how can we possibly justify writing $$\dfrac{d}{dx}=\dfrac{dy}{dx}\dfrac{d}{dy}?$$

One of the answers to the previous question mentioned that if

$$\frac{du}{dx}=\frac{du}{dy}\sqrt{\frac{m\,\omega_0}{\hbar}}$$ then just looking at the operator $\frac{d}{dx}$ without explicitly addressing the function $u$ the operator is to apply: \begin{align*} \frac{d}{dx}&=\sqrt{\frac{m\,\omega_0}{\hbar}}\frac{d}{dy} \end{align*}

In one of the comments given to the other answer to the previous question states that

I find it easier to look at what is $\dfrac{du}{dx}$ is in terms of $y$ this means working out the derivative of $u(y(x))$ to do this you use the chain rule - so we have $$\frac{du}{dx} =\frac{dy}{dx}\frac{du}{dy}$$ so looking at the operator part I.e "ignore" the $u$. This is how my physicist brain computes changes of variables.

But I am finding it hard to accept that we simply "ignore" the $u$. I realize that we cannot simply cancel out the $u$ since we are not guaranteed that $u\ne 0$.

Is there a more plausible explanation as to why we may write $$\dfrac{d}{dx}=\dfrac{dy}{dx}\dfrac{d}{dy}$$

in the absence of a function to operate upon?

Regards.

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    $\begingroup$ Personally, I don't like the $dx$, $dy$ notations in calculus. Because it is not clearly reflects what we are doing(such as what functions we differentiate with, or applied by the chain rule). In fact, either in calculus or analysis, we can always avoid such $d-$like notation, and use a more rigourous manner. $\endgroup$
    – Eric
    Commented Nov 13, 2016 at 0:34
  • $\begingroup$ @Eric I agree completely; What is this more rigorous manner you speak of? Many thanks. $\endgroup$
    – BLAZE
    Commented Nov 13, 2016 at 0:36
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    $\begingroup$ I suggest to read an analysis book, not a calculus one. Calculus books such like Stewart, Larson, Thomas, and so on, are all using informal and ill symbols and concept, that will definitely cause the perspective readers uncomfortable and queer. Among the analysis books, I highly suggest you The Real Number and Real Analysis, the author had discussed illness and abuse of notations in the elementary calculus books. $\endgroup$
    – Eric
    Commented Nov 13, 2016 at 0:42
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    $\begingroup$ From that book, you can know that every informal use of symbol that calculus books often take, such like treating chain rule like this ugly style: $\frac{du}{dx}=\frac{dy}{dx}\frac{du}{dy}$, and treat the inverse function theorem as $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$ like a fraction, can be avoided. $\endgroup$
    – Eric
    Commented Nov 13, 2016 at 0:44
  • $\begingroup$ @Eric, while I'm sure many books abuse the notation, are you aware that $\frac{\partial}{\partial x}$ has very precise meaning in differential geometry? Using $d$ instead of $\partial$ is common when talking about one-dimensional manifolds such as $\Bbb R$. On the other hand $dx$ has yet another, precise meaning in diferential geometry. $\endgroup$
    – Ennar
    Commented Nov 13, 2016 at 18:51

2 Answers 2

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If two operators are equal, it just means that if they operate on the same function, they generate the same result. So $$\dfrac{d}{dx}=\dfrac{dy}{dx}\dfrac{d}{dy}$$ just means:

For all functions $u$ for which both sides are defined, $\dfrac{du}{dx}=\dfrac{dy}{dx}\dfrac{du}{dy}$.

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  • $\begingroup$ Thanks for your answer. Which two operators are equal here? $\endgroup$
    – BLAZE
    Commented Nov 13, 2016 at 0:39
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    $\begingroup$ $\dfrac{d}{dx}$ and $\dfrac{dy}{dx}\dfrac{d}{dy}$. $\endgroup$
    – TonyK
    Commented Nov 13, 2016 at 0:43
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Here are some aspects regarding the shift from \begin{align*} \dfrac{du}{dx}=\dfrac{dy}{dx}\cdot\dfrac{du}{dy}\qquad\longrightarrow\qquad \dfrac{d}{dx}= \dfrac{dy}{dx}\cdot\dfrac{d}{dy}\tag{1} \end{align*}

We do not forget anything, but instead we change our point of view to a more abstract one.

First step: Functions of points

At first we take a look at a simpler example, one step below the abstraction layer in (1). We consider real-valued functions $f,g:\mathbb{R}\rightarrow \mathbb{R}$ and scalar multiplication. As we know \begin{align*} c\left(f(x)+g(x)\right)=cf(x)+cg(x)\qquad \forall x\in \mathbb{R}\tag{2} \end{align*}

We can express this relationship more abstract by writing \begin{align*} c(f+g)=cf+cg\tag{3} \end{align*}

We know functions transform real values to real values. But instead of applying functions to specific values as we did in (1) we now begin to treat them in (2) as objects by their own.

In (1) we consider scalar multiplication and addition of real values $c(f(x)+g(x))$. In (2) we consider scalar multiplication and addition of functions.

Of course, we don't forget that functions are objects which are applied to reals. But we shift our view to a more abstract one. This has a tremendous benefit. From now on we can ask where do functions live and how do they interact. In the same way as we did former ask, where do reals live and how do they interact.

We can consider the set of real-valued functions and study the relationship between elements of this set as we formerly did when we considered the set of reals and studied the relationship between them.

In fact this is the first step in a direction where functions become points in a function space and where we study the relationship of the points within such spaces. This is the main theme of functional analysis.

Second step: Functions of functions

We now consider real-valued differentiable functions $f,g$ in one variable. We look at the chain rule of differentiation \begin{align*} \frac{d}{dx}\left(f\left(g(x)\right)\right)=\frac{d}{dx} g(x)\cdot\frac{d}{dg}f(g)\tag{4} \end{align*} We can express this relationship more abstract by writing \begin{align*} \frac{d}{dx}=\frac{dg}{dx}\cdot\frac{d}{dg}\tag{5} \end{align*}

Note the step from (4) to (5) is analogously to the step from (2) to (3). In (4) we see a differential operator transforms a function to a function. In (5) we take a more abstract point of view and consider multiplication of differential operators. Here we are going towards operator calculus.

Notes:

  • Regarding some comments and a seemingly abuse of notation when using $\frac{dy}{dx}$.

    Keep in mind that this notation is also extremely powerful. It indicates interesting relationships only by its pure power of symbols which other notational conventions can't do.

    It can be made mathematically rigorous as it is shown in this answer.

  • See this paper for some information about how and when it's convenient to work with operators.

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  • $\begingroup$ In your answer on that post you included, you defined and clarify the meaning of $dy$, $dx$, the so called differential. But these $dy$ and $dx$ term are absolutely unrelated to those $dy$ and $dx$ that people write in $\frac{dy}{dx}$. The $dy$ and $dx$ of the differential meaning, is not the same as the same two symbol in $\frac{dy}{dx}$. (In fact, the differential symbol $dy,dx$ is quite natural and rigorous for me, but in the contrary, I personally consider $\frac{dy}{dx}$ to be an abuse of notation. BTW, I know that Robinson have tried to formalized them.) $\endgroup$
    – Eric
    Commented Nov 14, 2016 at 12:42
  • $\begingroup$ It is to say, it seemed to me that you didn't clarify the meaning of the $dy$, $dx$ symbol in $\frac{dy}{dx}$, or something like $\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}$. Instead, you just defined and talked on a literally whole different thing, making these clear and reasonable, but these, are not the original one that I discuss, or other people who used to feel the $dx$, $dy$ symbol weird curious about. $\endgroup$
    – Eric
    Commented Nov 14, 2016 at 12:43
  • $\begingroup$ @Eric: I don't think $dy$ and $dx$ are absolutely unrelated to $\frac{dy}{dx}$. I've added some info about the relationship which might help to clarify the situation. $\endgroup$ Commented Nov 15, 2016 at 11:06
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    $\begingroup$ @BLAZE: Some info added to the referenced answer regarding the differential quotient. $\endgroup$ Commented Nov 15, 2016 at 11:07

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