Is $\frac{\partial}{\partial x}(x-y)$ ill-defined?

Question

Let $f(x)=x-y$. Then $$\frac{\partial f}{\partial x}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\frac{x+h-y_h-x+y}{h}=\lim_{h\to 0}\frac{h+y-y_h}{h},$$ noting that $y$ can depend on $x$ (that's why I wrote $y_h$).

Now if instead we let $f(x,y)=x-y$, then $$\frac{\partial f}{\partial x}=\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}=\lim_{h\to 0}\frac{x+h-y-x+y}{h}=1$$ which is, in general, not equivalent the first result.

Does this mean that $\frac{\partial}{\partial x}(x-y)$ is an ill-defined expression?

Answer 1

The only way in which $\frac{\partial}{\partial x}(x-y)$ is ill-defined, is the same way in which the function $x-y$ is ill-defined. That is, the very instant we define the function $x-y$, its partials are also well-defined.

For instance:

• We define $f(x,y) = x-y$. Then immediately we'd have $\frac{\partial f}{\partial x} = 1.$ Note that this doesn't depend on whether $x,y$ are dependent in your application, as the function $f$ itself necessarily has no information about this dependency in order to define its partials.
• We define $g(x) = x - y(x)$. Then we'd have $\frac{\partial g}{\partial x} = 1 - y'(x)$.
• We define $G(x, t) = x-y(x,t)$. Then we'd have $\frac{\partial G}{\partial x} = 1 - \frac{\partial y}{\partial x}.$ (note here that the definition of $\frac{\partial y}{\partial x}$ requires fixing a function for $y$ in terms of $x,t$, as if we only mean $y$ depends on $x,t$, there isn't enough information to define $\frac{\partial y}{\partial x}$)
• In your instance of $h(x) = x-y$, this function $h$ is ill-defined, so its partials will be too.

How we define $\frac{\partial}{\partial x} (x-y)$ itself will be context-dependent.