I have the following differential form of a bivariate function $f(x,y)$
$$\text{d}f = A(x,y)\text{d}x+B(x,y)\text{d}y \hspace{1 in} \text{Eq. 1}$$
I wish to express this as a first order accurate approximation as follows, and my question is how to represent the order of the remainder (error) term
$$\Delta f = A(x_0,y_0)\Delta x+B(x_0,y_0)\Delta y + \text{O}(?) \hspace{1 in} \text{Eq. 2}$$
I was thinking of something like
$$\Delta f = A_0\Delta x+B_0\Delta y + \text{O}[\text{Max}(|\Delta x|^2,|\Delta y|^2)] \hspace{1 in} \text{Eq. 3}$$
or
$$\Delta f = A_0\Delta x+B_0\Delta y + \text{O}[\text{Max}(|\Delta x|,|\Delta y|)^2] \hspace{1 in} \text{Eq. 4}$$
but was wondering if there was a more compact way of representing this.
Note: $$ A_0=A(x_0,y_0)\\ B_0=B(x_0,y_0) $$
The motivation behind the finite difference approximation above is that I ultimately wish to divide Eq. 2 by $\Delta x$ and take the limit as $\Delta x \rightarrow 0$, (noting that in this limit, $\Delta y$ must $\rightarrow 0$ for the limit to exist) so that I have
$$ \frac{\partial f}{\partial x}\Bigg|_{x_0,y_0} =A_0 +B_0 \frac{\partial y}{\partial x}\Bigg|_{x_0,y_0} \hspace{1 in} \text{Eq. 5} $$
I believe this is the more rigorous way of proving Eq. 5 as opposed to dividing Eq. 1 by the differential $\text{d} x$ and assuming that the ratio of two differentials is a partial derivative, even if it does yield the correct expression.
Thoughts? Thanks.