How to take derivatives if we re-write a function $f(x,y)$ as $f(x/y)$, $f(x-y)$, etc

Question

For example, consider $f(x,y)=x/y$

Can we instead write this as $f(x/y)=x/y$. Is this valid?

I think it is if we consider it a change of variables (i.e. $u=x/y$ and $g(u)=u)

But what if we do not change the variable and just leave it as $f(x/y)$? Can we even take the partial derivative of this w.r.t $x$? If so how (Would we still use the chain rule?)

A similar example would be writing $f(x,y)=x-y$ as $f(x-y)$

Answer 1

1. If you as $f(\frac{x}{y})$ that is some sort of notational abuse, doesn't make much sense to me.
2. Yes, now $u$ is a function of $(x,y)$
3. If you have a function in the arguments like something of the form: $$ T(x,y) = F(x-y)$$

Then you can take the partial derivative using chain rule.

$$ \frac{\partial T}{\partial x} = \frac{\partial F}{\partial u}|_{x-y} \frac{ d(x-y)}{d x}$$

Of course, the story is different if $y$ is a function of $x$ then you'd have to use total derivatives.

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We can see that $g$ is dependent on $u$ and $u$ is dependent on $(x,y)$. If we are take the derivative of $g$ with $u$ directly, we don't care what caused $u$ to change, we just care of how $g$ would change if some how we could nudge the $u \to u + du$. Hence,

$$ \frac{dg}{du} = 1$$

While say if you took the partial derivative of expression with $x$ that is:

$$ G(x,y) = g(u(x,y))$$

And then took a partial,

$$ \frac{ \partial G}{\partial x} = \frac{ \partial g}{\partial u} \frac{du}{dx}$$

This we can think of 'breaking the connection' between $y$ and $u$/ keeping $y$ as fixed and seeing how the function changes as a whole as we crank $x$