Partial Derivative Chain Rule When Variables Are Not Independent 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?
 A: 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.
