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It seems clear enough to define the total derivative of a function $f=f(x(t),y(t))$ of multiple "intermediate variables" $x$ and $y$, who themselves depend on one independent variable $t$, to be:

$$\frac{df}{dt}=\frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$$

(One could say this is simply an application of the multivariable chain rule.)

I also find it reasonable when multiple independent variables $t$ and $u$ are present (yielding $f=f(x(t,u),y(t,u))$ to calculate the partial derivatives by:

$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}$$

(similarly for $\frac{\partial f}{\partial u}$.)

However, I am confused if such as thing as the "total derivative" $\frac{d f}{d t}$ can be defined when their are multiple independent variables $t$ and $u$ present, and how it would relate to the partial derivative $\frac{\partial f}{\partial t}$.


Note: I understand that in the simple case of a function depending explicitly on multiple independent variables, such as $f=f(t,u)$, we do have total derivatives with respect to each variable:

$$\frac{df}{dt}=\frac{\partial f}{\partial t} + \frac{\partial f}{\partial u} \frac{du}{dt} \text{ and } \frac{df}{du}=\frac{\partial f}{\partial t} \frac{dt}{du} + \frac{\partial f}{\partial u}$$ I'm not sure how this can be generalized to the case with $f$ depending on intermediate variables $x(t,u)$ and $y(t,u)$.


I attempted to solve this myself, but I ended up with a strange result. Here was my approach:

Take the total differential of $f$:

$$df=\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$

being aware that since $x=x(t,u)$, then $dx=\frac{\partial x}{\partial t} dt + \frac{\partial x}{\partial u} du$ and similarly for $dy$.

Dividing the total differential by $dt$ (not rigorous, I know), we obtain:

$$\frac{df}{dt}=\frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$$

We can evaluate $\frac{dx}{dt}$ as $\frac{dx}{dt}= \frac{\partial x}{\partial t} \frac{dt}{dt} + \frac{\partial x}{\partial u} \frac{du}{dt} = \frac{\partial x}{\partial t}$ because $\frac{du}{dt}=0$, and similarly for $\frac{dy}{dt}$:

$$\frac{df}{dt}=\frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}$$

But I must have made a mistake somewhere, because this expression is equal to $\frac{\partial f}{\partial t}$, which should only be true when $t$ is the only independent variable. What have I done wrong, and what is the true expression for $\frac{df}{dt}$ (if it exists)?

Thank you for your time.

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  • $\begingroup$ The definition of the derivative is that the derivative is a linear transformation. Spivak's Calculus on Manifolds is an excellent reference. $\endgroup$ May 8 '21 at 0:20
  • $\begingroup$ I'm sorry, I don't see how that helps discern whether there is a difference between "total" and "partial" derivatives for a function of multiple implicit variables. $\endgroup$ May 8 '21 at 1:36
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    $\begingroup$ I don't see why we have the term "total derivative" at all. The only concept we need is the "derivative" of a function $f: \mathbb R^n \to \mathbb R^m$. Partial derivatives are ok too. For example, if $\hat f(t) = f(x(t), y(t))$, then the derivative of $\hat f$ is $\hat f'(t) = D_1 f(x(t),y(t)) x'(t) + D_2 f(x(t), y(t)) y'(t)$. That's just the derivative of $\hat f$, not the "total derivative". The term "total derivative" seems to add nothing and only cause confusion. $\endgroup$
    – littleO
    May 8 '21 at 2:38
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    $\begingroup$ Another common source of confusion is that in the equation $\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$, two different functions are being called by the same name $f$. We should give the function $f(x(t), y(t))$ its own name, because it's not the same function as $f$. $\endgroup$
    – littleO
    May 8 '21 at 2:41
  • $\begingroup$ @littleO I see, thank you $\endgroup$ May 10 '21 at 13:01
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The problem is not writing compositions properly. If $f$ is a function of $x$ and $y$ and those are functions of $t$ and $u$, then $t \mapsto f(x(t,u),y(t,u))$ is a function of $t$ and $u$, but $f$ itself is not a function of $t$ and $u$, this is an abuse of notation! Rigorously, one defines $\widetilde{f}(t,u)=f(x(t,u),y(t,u))$ and the chain rule says that $$\frac{\partial \widetilde{f}}{\partial t}(t,u) = \frac{\partial f}{\partial x}(x(t,u),y(t,u)) \frac{\partial x}{\partial t}(t,u)+\frac{\partial f}{\partial y}(x(t,u),y(t,u)) \frac{\partial y}{\partial t}(t,u).$$Introducing this extra letter $\widetilde{f}$ just for the sake of being formal is something that people usually don't try to do, so what you're denoting by $\partial f/\partial u$ is actually $\partial \widetilde{f}/\partial u$ for $\widetilde{f}$ defined as above.

If $x$ (for instance) is a function of $t$ and $u$, ${\rm d}x/{\rm d}t$ doesn't make sense, only $\partial x/\partial t$ does. One can make sense of this using the same mechanism above. Namely, define $\widetilde{x}(t) = x(t,u)$ where $u$ is fixed. So $$\frac{{\rm d}\widetilde{x}}{{\rm d}t}(t) = \frac{\partial x}{\partial t}(t,u),$$and $u$ is fixed everywhere throughout this process. One could even denote $\widetilde{x}$ by $\widetilde{x}_u$ (not a partial derivative) to make explicit that a given $u$ has been chosen.

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    $\begingroup$ Good answer. Not giving $\tilde f$ its own name causes a lot of confusion in my opinion. $\endgroup$
    – littleO
    May 8 '21 at 2:44
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    $\begingroup$ Perhaps one reason for not giving it its own name is related to how in physics the name of the function is often the same as the variable for some quantity, like $x$ for position, or $t$ for time. Mathematicians might prefer that we use an $\tilde{x}$ if we introduce a new function (e.g. with a different domain), but in physics I think it would be confusing to have different symbols that actually represent the same quantities (even if they are different functions). $\endgroup$
    – Joe
    May 8 '21 at 11:45
  • $\begingroup$ Great answer, thank you. $\endgroup$ May 10 '21 at 12:59
  • $\begingroup$ To summarize your argument, would it be correct to say that a function of multiple "intermediate variables" which are functions of multiple independent variables cannot have a "total" derivative, only partial derivatives? That is, for $f\tilde (t,u) \equiv f(x(t,u),y(t,u)$, there is no good sense in which to define $d f\tilde/dt$, only $\partial f\tilde/\partial t$? $\endgroup$ May 12 '21 at 1:01
  • $\begingroup$ No, total derivatives always make sense, I just didn't mention it because it wasn't directly relevant to addressing what I perceived as your confusion. As someone else said in the comments, see Spivak's Calculus on Manifolds for more on that. $\endgroup$
    – Ivo Terek
    May 12 '21 at 1:42

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