0

I was wondering if someone knows what exactly the error is in the following manipulations involving partial derivatives. Supposing that $y=x^2$ then $\frac{\partial}{\partial x} \frac{\partial}{\partial y} y = \frac{\partial}{\partial x} 1 = 0$ but $\frac{\partial}{\partial y}\frac{\partial}{\partial x} y = \frac{\partial}{\partial y} 2x = \frac{\partial}{\partial y} 2 y^{1/2} = y^{-1/2}$. I feel like it must be something to do with treating y as an independent variable but I'm not sure what the rigorous reason is?

A quarky name
  • 25
  • Hi, I hope this is clearer. Essentially $\partial_y = \frac{\partial}{\partial y}$ for instance. – A quarky name Jul 30 '21 at 20:11
  • 1
    That is only true when x and y are independent variables. Here y is dependent on x. Hence, you can't commute the operators. – Harshith vallabhaneni Jul 30 '21 at 20:16
  • 2
    What you are doing is not partial differentiation. Not even remotely close to it. – David K Jul 30 '21 at 20:18
  • To have partial derivatives you need a function of $n$ inputs, and the partial derivative is what happens if you vary one input while holding all other inputs constant. It has nothing to do with the names of variables; the $\partial_x$ notation is merely a convenient way to identify which input you're varying when you have labeled all the inputs. Here you have two variables but you have not identified them with the inputs of any two-variable function. – David K Jul 30 '21 at 20:23
  • 1
    Hi, thanks a lot for your comments. What I wrote did seem very illegitimate. Essentially the question arose when considering another function $g=g(y(x),z)$ and considered differentiating first with respect to $y$ and then with respect to $x$. Is this a valid thing to consider doing? Perhaps this would require taking the total derivative w.r.t x after the partial w.r.t $y$? – A quarky name Jul 30 '21 at 20:30
  • First, I do not see a clear way (without much further clarification of the problem) that you can take any kind of derivative of $g(y(x),z)$; second, why do you want to? And why differentiate twice? – David K Jul 31 '21 at 22:52
  • Hi, so basically this is coming up in a physics context where I have the function \begin{equation} F(g(x_1(s),x_2(s),...,x_n(s)),x_1(s),x_2(s),...,x_n(s), y,z) \end{equation} – A quarky name Aug 01 '21 at 12:01
  • In order to derive dynamical equations I believe that I need to evaluate $\frac{\partial }{\partial x_i} \frac{\partial F}{\partial g}$. I'm getting confused by the fact that it seems that the $x_j$ (at least in the right combination) seem to depend on $g$ (say $g=x_1+x_2$ and $F$ contains $x_1$ and $x_2$ in the combination $x_1+x_2$ (for example I wouldn't think $ \frac{\partial x_1}{\partial g} = \frac{\partial x_2}{\partial g} = 0$ like they are independent as then $\frac{\partial (x_1+x_2)}{\partial g} = \frac{\partial g}{\partial g} = 0$ seemingly? – A quarky name Aug 01 '21 at 12:01
  • I do agree that we shouldn't expect these partials to commute however. – A quarky name Aug 01 '21 at 12:01
  • Partials that don't commute would still lead me to suspect that they're being interpreted incorrectly. At the very least, lack of commutativity implies that the partial derivatives are discontinuous (if they even exist). – David K Aug 01 '21 at 20:49
  • Mathematically, $\frac{\partial }{\partial x_i}\frac{\partial}{\partial g} F(g(x_1(s),x_2(s),...,x_n(s)),x_1(s),x_2(s),...,x_n(s), y,z)$ is simply nonsense, or at best an egregious abuse of notation. Even taking a partial derivative with respect to $x_i$ is questionable due to the common argument $s$ in all the $x_i$ arguments. But convenience wins out over unambiguous mathematical expression in some practical applications. There's a related discussion in an answer to another question that I think is is worth understanding. – David K Aug 01 '21 at 20:49
  • Ultimately, since the partial derivative comes from a physics problem, I think the only way to understand it correctly is through the physics that it represents. That means not only knowledge of the specific physics phenomena represented by your functions but also the specific purpose of taking multiple partial derivatives. Altogether this seems like a question totally unsuited to math.SE; you might do better by taking the actual physics problem (not abstracted at all!) to physics.SE where you're likely to find people who know what these derivatives are really supposed to do. – David K Aug 01 '21 at 20:50
  • It might have been useful to know, before writing the comments above, that you had already continued this line of inquiry in Partial derivative with respect to function of a variable and Confusion about differentiating a variable with respect to a function of that variable (and others). I wonder if you got any enlightenment from the responses to any of the three questions. I also wonder why you are so sure that mathematicians rather than physicists are the people to answer a physics problem. – David K Aug 05 '21 at 03:42
  • Hi, thank you again for your comments and links (which were also rather helpful in thinking about the issue of notation). Apologies for not linking my other questions - I didn't want to spam this comment section and I am still rather new to the site so am not sure what protocol is in this regard. I do believe I have a better understanding of where my conceptual difficulties lie now. I think I felt that the precision of mathematicians would be helpful in sorting out what seemed to be an issue of correct definitions (ultimately this does appear to be the root of my misunderstanding). Cheers! – A quarky name Aug 05 '21 at 10:16

1 Answers1

0

Your $y$ is a function of the variable $x$. Hence $\frac{\partial}{\partial y}y =1 $ is wrong, because $y$, being a function of $x$, does not depend on $y$ at all. For all purposes, I can replace your $y$ by $f(x)$, and then obviously $\frac{\partial}{\partial y}f =0$, and so is $\frac{\partial}{\partial f}f$, which may look a bit unusual, and for a good reason too - it is unusual because it is nonsensical.

uniquesolution
  • 18,541
  • Hi, thank you for your answer. I was wondering if I could ask a follow-up question. Supposing that we had the function $f=f(y(x))=y(x)= x^2$ what would be the error in the manipulations: $\frac{\partial}{\partial x} \frac{\partial}{\partial y} f = 0$ but $\frac{\partial}{\partial y} \frac{\partial}{\partial x} f = \frac{\partial}{\partial y} (2x) = 2/(\frac{\partial}{\partial x}y)=1/x$? – A quarky name Jul 31 '21 at 08:11