Is the total derivative defined for a function of multiple independent variables?

Question

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 $\frac{\partial f}{\partial t}$.

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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 $y(t,u)$.

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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.

Answer 1

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.