Partial derivatives vs. Total Derivatives for chain rule. If I had a function $f(x,y)$ where $x=x(s,t)$ and $y=y(s,t)$ then 
$$ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}. $$
I have been reading this and have spotted in case 2 that 
$$ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}. $$
$f$ has no explicit $t$ dependence so how can this be correct? How can I have a chain rule for partial derivatives? A chain rule implies that there is no explicit dependence on the variable that we are differentiating with respect to, and I thought partial derivatives only deal with explicit dependencies and not implicit? Thanks
 A: In case $2$, it's not $\frac{\partial f}{\partial t}$, it's $\frac{\partial z}{\partial t}$, and $z$ is not independent of $t$.
A: In a strict mathematical sense, you are correct. BUT: Engineers, physicists and other people who depend highly on calculus depend to simplify things, or drop parts of the standard notation.
In this case they mean $$\frac{\partial f(x,y)}{\partial t}(s,t) = \frac{\partial f}{\partial x}(x(s,t),y(s,t)) \frac{\partial x}{\partial t}(s,t) +  \frac{\partial f}{\partial y}(x(s,t),y(s,t)) \frac{\partial y}{\partial t}(s,t) $$
As you can see, this is quite a lot to write (I hope, I did no typo's) Mathematicians tend to drop the $(s,t)$ and then do not mean that the evaluations of the functions are the same, but the functions are the same. Physicists (and other people mentioned obove) tend to even drop the dependency of $f$ and by that kind of overwrite the meaning of the old f by the composition with the new f....
A: An ordinary derivative--such as that of $f$ with respect to $t$, in your example is only valid if all the dependencies of $f$, all the dependencies of those dependencies, all the dependencies of those dependencies, and so on, can be shown to have a single common dependency $t$, explicit or otherwise. In all other cases, only a partial derivative applies, since $f$ then only depends partially on $t$. 
That is why a partial derivative is used in Case 2; $t$ is not a common dependency of the system $(f,x,y,s,t)$, as $s$ is in no way dependent upon $t$.
