Prove $ \Delta z = f(x+\Delta x,y+\Delta y) - f(x,y) = f_x \Delta x + f_y \Delta y + \alpha\Delta x+ \beta\Delta y $ is $f(x,y)$ is differentiable I need to prove the following useful statement:
If $f(x,y)$ is differentiable at $(x_0,y_0)$, then in the neighbourhood of $(x_0,y_0)$, we have
$$
\Delta z = f(x_0+\Delta x,y_0+\Delta y) - f(x_0,y_0) = f_x \Delta x + f_y \Delta y + \alpha\Delta x+ \beta\Delta y
$$
where $\alpha \rightarrow 0, \beta \rightarrow 0$ when $\rho = \sqrt{(\Delta x)^2+(\Delta y)^2} \rightarrow 0$.
My try: according to the definition of differentiability, we can write
$$
\Delta z = f(x_0+\Delta x,y_0+\Delta y) - f(x_0,y_0) = f_x \Delta x + f_y \Delta y + o(\rho)
$$
Here comes the vague thing, how to get $\alpha\Delta x+ \beta\Delta y$ from $o(\rho)$?
Any thoughts? Thanks!
 A: Hope you are aware of little $o$ and big $O$. I assume it is a little $o$. Using Taylor series expansion, it will be $O(\Delta x^2 + \Delta y^2)$, which can be eventually written as $o(\Delta x+ \Delta y)$ by the definition.
A: For reference: Dr. Wilfred Kaplan takes it as a definition for differentiability in his book "Advanced Calculus", on pp86 eq(2.16-2.17). It states as follows,

In general, the function $z=f(x, y)$ is said to have a total differential or to be differentiable at the point $(x, y)$ if, at this point,
$$
\Delta z=a \Delta x+b \Delta y+\epsilon_{1} \cdot \Delta x+\epsilon_{2} \cdot \Delta y
$$
where $a$ and $b$ are independent of $\Delta x, \Delta y$ and $\epsilon_{1}$ and $\epsilon_{2}$ are functions of $\Delta x$ and $\Delta y$ such that
$$
\lim _{\substack{\Delta x \rightarrow 0 \\ \Delta y \rightarrow 0}} \epsilon_{1}=0 . \quad \lim _{\substack{\Delta x \rightarrow 0 \\ \Delta y \rightarrow 0}} \epsilon_{2}=0
$$

The problem still remains, how to prove the equivalence of these two definitions for differentiability? And in what sense should we define the equivalence? For convenience, the other one def is as follows,
$$
\Delta z=a \Delta x+b \Delta y + o(\rho)
$$
where $\rho = \sqrt{(\Delta x)^2 +(\Delta y)^2}$ and
$$
\lim_{\rho\rightarrow 0} \frac{o(\rho)}{\rho} = 0
$$
A: Finally, I developed a proof, please check on this.

Basically, this question is about the equivalence of these two definitions of differentiability, which further reduces to this question: prove that $o(\rho)$ equivalent to $\alpha \Delta x + \beta \Delta y$. To prove this, we need to prove in both directions: 
(1) $o(\rho)$ can be written in the form of $\alpha \Delta x + \beta \Delta y$. 
(2) $\alpha \Delta x + \beta \Delta y$ is actually an $o(\rho)$.
The proof of the second one is trivial. Let's focus on the first one. 
Proof of (1):
Since
$$
\lim_{\rho \rightarrow 0} \frac{o(\rho)}{\rho} = 0
$$
we have
$$
o(\rho) = \gamma \rho = \gamma \sqrt{(\Delta x)^2+(\Delta y)^2}
$$
where $\lim_{\rho\rightarrow0} \gamma = 0$. Further we have
$$
o(\rho) = \frac{\gamma\rho}{a |\Delta x| + b |\Delta y|}(a |\Delta x| + b |\Delta y|) = \frac{\gamma\rho}{a |\Delta x| + b |\Delta y|}a |\Delta x| + \frac{\gamma\rho}{a |\Delta x| + b |\Delta y|}b |\Delta y| 
$$
where $a,b$ are constants and we assume $0<a\leq b$ without loss of generality. Let
$$
\alpha = \frac{\gamma\rho}{a |\Delta x| + b |\Delta y|}a, \beta = \frac{\gamma\rho}{a |\Delta x| + b |\Delta y|}b
$$
Now we need to prove $\lim_{\rho \rightarrow 0} \alpha = 0, \lim_{\rho \rightarrow 0} \beta = 0$. Since
$$
|\alpha| = \left|\frac{\gamma\rho}{a |\Delta x| + b |\Delta y|}a \right| \leq \frac{|\gamma| a(|\Delta x| + |\Delta y|)}{a(|\Delta x| + |\Delta y|)} = |\gamma| \rightarrow 0
$$
Therefore, $\lim_{\rho \rightarrow 0} \alpha = 0$. So the same with $\beta$. 
To conclude, we have
$$
o(\rho) = \alpha |\Delta x| + \beta |\Delta y| 
$$
where $\lim_{\rho \rightarrow 0} \alpha = 0, \lim_{\rho \rightarrow 0} \beta = 0$. The absolute value operator can be readily removed since it can be included in $\alpha$ and $\beta$. Thus we have
$$
o(\rho) = \alpha \Delta x + \beta \Delta y
$$
Statement (1) is proved. The equivalence between $o(\rho)$ and $\alpha \Delta x + \beta \Delta y$ is established. So the same with the equivalence of the two definitions of differentiability.
