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Similar post, but opposite conclusion?

I have a function $x(t)$ and its derivative as a function of other variables$\frac{dx}{dt}(x(t))$ (a differential equation?) and I want to find $\frac{\partial }{\partial x}\frac{dx}{dt}$. The linked post above says this is zero, and I can see where they are coming from swapping the order of the derivatives. However, I just don't believe it, I have likely misunderstood their point. For instance, say $\frac{dx}{dt}=x^2$. This is easily solvable, e.g. $x=\frac{-1}{t}$. But $\frac{\partial}{\partial x}\frac{dx}{dt}=2x$ which is not zero in general. Yet $\frac{\partial}{\partial x}\frac{d}{dt}x=\frac{d}{dt}\frac{\partial}{\partial x}x=\frac{d}{dt}1=0$? What goes wrong here?

EDIT: is it because $x$ is a function of $t$? So we cant swap them?

user947124
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  • It seems you found a simple counterexample to both answers in the other post. How about continuing with the discussion there? I guess it is worth clarifing a few things in one thread. – Kurt G. Jul 23 '21 at 12:36

1 Answers1

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It doesn't really make sense to talk about the operation you're describing.

On one hand, you have a function $f$, say $f(x)=x^2$, which can be differentiated, giving $f'(x)=2x$. Here the argument happens to be named $x$, so we can say that we are “differentiating $f(x)$ with respect to $x$”, but we could equally well say that if $f$ is given by the expression $f(y)=y^2$, then its derivative is given by the expression $f(z)=2z$ (for example). So it's clearer to just say that the derivative of the function $f$ is the function $f'$.

On the other hand, you have a differential equation $x'(t)=f(x(t))=x(t)^2$, whose solution is a function $x(t)$, which also can be differentiated. Its argument happens to be called $t$ here, so it makes sense to say “differentiate $x(t)$ with respect to $t$”, but not “with respect to $x$” or “with respect to $x(t)$”.

Hans Lundmark
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  • I suppose, in particular, it does not make sense to swap the derivatives in $\frac{\partial}{\partial x}\frac{dx}{dt}$ which was a line of argument in that other thread. Right? – Kurt G. Jul 23 '21 at 12:50
  • Could you elaborate on how you can differentiate the function $x\mapsto x^2$ with respect to $x$? In my experience, the phrase "differentiate with respect to" is often used in the context of differentiating an expression like $x^2+t$. In this context, differentiating $x^2+t$ with respect to $x$ means (i) identity the expression $x^2+t$ with the function $f$ such that, for all $x$, $f(x)=x^2+t$, and (ii) evaluate $f'$ at $x$. – Joe Jul 23 '21 at 12:54
  • @Joe: I agree with what you're writing, but I'm not sure I understand what your point is, or what I'm supposed to elaborate. I personally find it clearer to just talk about $f$ and $f'$, without worrying about what the name of the variable might be, but here I was trying to connect to the “with respect to” terminology used in the question (which I think is the source of the whole confusion). – Hans Lundmark Jul 23 '21 at 13:11
  • @KurtG.: The question in that thread makes even less sense to me, so I don't think I have anything intelligent to say about that. – Hans Lundmark Jul 23 '21 at 13:13
  • @Joe: I reprased the explanation a little. Maybe it's clearer now? – Hans Lundmark Jul 23 '21 at 13:25
  • @HansLundmark: My point was that the phrase "differentiate with respect to" is only used as a disambiguation tool: if I say "differentiate $x^2+t$ with respect to $x$", I'm letting the reader know that I am interested in computing $(x\mapsto x^2+t)'(x)$ rather than $(t\mapsto x^2+t)'(t)$. The expression $x^2+t$ does not clearly identity a function. – Joe Jul 23 '21 at 13:27
  • @HansLundmark: On the other hand, if I know for a fact that $f(x)=x^2$ or $f(x)=x^2+t$ or whatever, then the derivative of $f$ just means $f'$, and there isn't any ambiguity. Saying the derivative of $f$ with respect to $x$ is confusing in my opinion because functions just have derivatives—there aren't any "variables" involved. Of course, you know this, but I just consider this terminology a little bit confusing. By the derivative of $f$ with respect to $x$, do you just mean $f'(x)$? – Joe Jul 23 '21 at 13:28
  • @Joe: OK, then we agree completely! I also think this terminology causes much confusion. And yes, I just mean $f'(x)$, so of course it's redundant to say “with respect to $x$”. But it still makes sense, which I don't think “the derivative of $f(x)$ with respect to $f$” does. – Hans Lundmark Jul 23 '21 at 13:32
  • I understand your point about labelling the variables, but I don't see the connection. What is the difference between $\frac{d}{dx}(x)$ and $\frac{\partial}{\partial x} x(t)$ ? I'm sure you would agree the former equals 1. I don't see why we would care about $x$ being a function of something else if we are taking the partial derivative. – user947124 Jul 23 '21 at 13:56
  • You can take the ordinary derivative of a single-variable function, or a partial derivative of a multi-variable function with respect to some of its arguments. But you can't take the derivative of a function with respect to something which is not one of its argument, since the derivative is about varying the argument and seeing how the function responds. – Hans Lundmark Jul 23 '21 at 15:35
  • When you write $x(t)$, $x$ is the name of the function and $t$ is the name of the argument, so the only derivative that makes sense is $x'(t) = \frac{d}{dt} x(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$. That is, you vary the input $t$ and see how the output $x(t)$ responds. You can't vary $x$ as an independent variable in this case, since $x$ denotes a function (a “dependent variable”), so it doesn't make sense to ask for the derivative of $x(t)$ with respect to $x$. – Hans Lundmark Jul 23 '21 at 15:35
  • But when you write $\frac{d}{dx}x$, the symbol $x$ plays a completely different role; it's the (independent) variable in an expression that describes an unnamed function. If you would name that function $f$, in other words $f(x)=x$, then the derivative could be written as $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = 1$. – Hans Lundmark Jul 23 '21 at 15:36
  • Just a light-hearted joke to lighten up the mood – Maximilian Janisch Jul 23 '21 at 18:05