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Let's say, $x$ is a function of $t$ ($x = x(t)$) and $y$ is a function of $t$ ($y = y(t)$). And, $f$ is a function of $x$ and $y$ ($f = f(x, y)$). Then by the chain rule $$\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$$. However, when you take the partial derivative of a function wrt a variable you keep all other variables constant. So, I am not sure how $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ above can be calculated, since you cannot keep $y$ constant if $x$ is allowed to be varied and vice-versa. Can someone please explain what I am understanding wrong?

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  • $\begingroup$ $\frac{\partial f}{\partial x}$ is calculated for $x$ independent variable and then it can be considered at point $(x(t), y(t))$. $\endgroup$
    – zkutch
    Commented Apr 12, 2022 at 1:58
  • $\begingroup$ $f(x,y)$ is a mapping from $\mathbb R^2 \rightarrow \mathbb R^1$, and its partial derivatives are computed independently of any other mappings, which in your case is $\mathbb R^1 \rightarrow \mathbb R^2$ and defined by functions $x(t),y(t)$. $\endgroup$
    – blamocur
    Commented Apr 12, 2022 at 2:28
  • $\begingroup$ To be clear, the chain rule generally is applied when the variables are mutually dependent on one another. $\endgroup$
    – ryang
    Commented Apr 12, 2022 at 4:29
  • $\begingroup$ I suggest you read this answer of mine and the various link in there. The issue is you need to be crystal clear on what the notation means and what it doesn't mean. $\endgroup$
    – peek-a-boo
    Commented Apr 12, 2022 at 5:10

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The issue is that you are using $x$ as both as a free variable, and as a function, $x(t)$. And likewise $y$.


The notation $\dfrac{\partial f}{\partial t}$ is really being used as an implicit shorthand for $\dfrac{\partial f(x(t),y(t))}{\partial t}$.

$~$ Where $f(x(t),y(t))$ is a convolution of the bivariate function $f$, with monovariate functions $x$ and $y$, each evaluated with the same argument $t$, a free variable.


Likewise you are using the notation $\dfrac{\partial f}{\partial x}$ is used as shorthand for $\left.\dfrac{\partial f(u,v)}{\partial u}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}$ .

$~$ That means to evaluate the partial differential of $f$ with respect to its first argument and then form a composition by substituting those arguments with $x(t)$ and $y(t)$.

$~$ Thus you are evaluating the partial differential of the field $f(u,v)$ over the $t$ parametised curve $\{\langle x(t),y(t)\rangle\}$


So the chain rule is actually:

$$\dfrac{\partial f(x(t),y(t))}{\partial t}=\left.\dfrac{\partial f(u,v)}{\partial u}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}\cdotp\dfrac{\partial x(t)}{\partial t}+\left.\dfrac{\partial f(u,v)}{\partial v}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}\cdotp\dfrac{\partial y(t)}{\partial t}$$

But this is annoying to read and write, so the shorthand is used for convenience.

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  • $\begingroup$ That makes sense. Thanks! $\endgroup$
    – dotseveral
    Commented Apr 12, 2022 at 12:38

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