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I have a function $f$ of one variable $t$, but I will write it in a funny way: using a second function $g$ as an intermediary: $$ f(g(t),t)=tg(t) $$ where $$ g(t)=t $$ . The exact definitions of $g$ and $f$ don't matter too much, anything will do the trick. Before I proceed, note that I (knowing the definition of $g$) could write $f$ in several ways (dropping the parentheses): $$ f=tg\qquad f=t^2\qquad f=g^2\qquad f=t^3/g $$ you can see that this could go on forever.

The issue arrives when I want to take partial derivatives. Going in the same order as before: $$ \frac{\partial f}{\partial t}=g\qquad \frac{\partial f}{\partial t}=2t\qquad \frac{\partial f}{\partial t}=0\qquad \frac{\partial f}{\partial t}=3t^2/g $$ and I could do the same with partials w.r.t $g$.

Now I recognize that because $f$ is only a function of $t$, I should even be taking partials with respect to it, but by the way I defined $f$ using $g$, the multivariable chain rule: $$ \frac{df}{dt}=\frac{\partial f}{\partial g}\frac{dg}{dt}+\frac{\partial f}{\partial t} $$ still requires a definition of the partial w.r.t. $t$. It should be noted that the total derivative w.r.t. $t$ (which should be $2t$ as $f(t)=t^2$) is retrieved from the multivariable chain rule if we keep the definitions of $f$ as a function of $g$ and $t$ consistent across the equation, i.e. if we just pick a definition and stick to it, it doesn't matter one, the total derivative will work.

What's going on here, all I am doing is variable manipulations, but somehow the calculus seems intrinsically tied to the particular definitions of $f$ in terms of the dependent variable that I just made up. In a sense that is obvious. But Still.

Am I doing something I am not allowed to do. Am I miscalculating something. Am I misinterpreting something. Obviously the partials aren't well defined if the variables are not independent. But there is more to it than that.

Though this came up in the context of Lagrangian mechanics, where we regularly evaluate partials w.r.t. "functions" that solely depend on $t$ (I suspect there is something variation-y about that stuff though), the problem is easy to state, only relying on beginner-level calculus, and has me stumped. Any help is appreciated :)

Arthur
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Tanner Legvold
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  • Here you are assuming $g$ is independent of $t$, which is not the case here. – Offlaw Feb 12 '20 at 08:10
  • @Offlaw How am I assuming that? – Tanner Legvold Feb 12 '20 at 08:11
  • Please try to compute partial derivatives from definition, then you will get my point. – Offlaw Feb 12 '20 at 08:17
  • Ok, I understand what you are saying. But I thought it was the point of a partial to pretend all the other variables are constant when you evaluate it. Even if they do depend on each other. I know I have seen examples in books where we have $L(\phi (t),\dot{\phi} (t),t)$ and then take partials w.r.t. $t$, $\phi$, or $\dot{\phi}$ without considering their interdependence. I'm sorry I can't give exact references right now, but it is all over Lagrangian mechanics. – Tanner Legvold Feb 12 '20 at 08:38
  • ahhh lagrangian mechanics, where the answers are easy to get, but understanding the meaning of intermediate steps is almost impossible without a lot of guidance (that's my personal experience). The real issue is the misleading use of the partial derivative notation, and not being careful with the distinction of a function (including its domain and target space) vs where it is being evaluated. at the moment I don't have time to elaborate, but I might later in the week (if you haven't already accepted an answer) – peek-a-boo Feb 13 '20 at 12:15
  • Its a year later, I found a helpful explanation of the Euler-Lagrange equation (which is was I was getting at) here – Tanner Legvold Mar 30 '21 at 20:55

1 Answers1

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You can write

$$f(t,g)=tg\implies \frac{\partial f}{\partial t}=g,$$ where $g$ is independent of $t$.

Or, with $g(t)=t$,

$$f(t,g(t))=tg(t)\implies\frac{df}{dt}= \frac{\partial f}{\partial t}+ \frac{\partial f}{\partial g}\frac{dg}{dt}=g(t)+t\frac{dg}{dt}=2t.$$

On the opposite,

$$\frac{\partial( t g(t))}{\partial t}=g(t)$$ is wrong (and using a partial derivative is ambiguous).

It is much safer to avoid mixing the formal and actual arguments, and write

$$f_u(u,v):=\frac{\partial f(u,v)}{\partial u},f_v(u,v):=\frac{\partial f(u,v)}{\partial v},$$

so that $$\frac{df(t,g(t))}{dt}=f_u(t,g(t))+f_v(t,g(t))\frac{dg(t)}{dt}.$$