f(x,y) jointly differentiable What is the definition of "jointly continuously differentiable function"? I.e. when $f:\mathbb{R}^2\rightarrow \mathbb{R}:(x,y)\mapsto z$ is jointly continuously  differentiable in $x,y$ ?
Is it $$\exists f'(x,y):=\lim_{(h_x,h_y)\rightarrow 0} \frac{f(x+h_x, y+h_y)-f(x,y)-ah_x -bh_y}{\sqrt{h_x^2+h_y^2}}\;\;\;\;(*)$$ s.t. $f'$ is continuous as a function $\mathbb{R}^2\rightarrow\mathbb{R}$?
($(*)$ edited after TIMH's answer)
 A: 
Is it $\exists f'(x,y):=\lim_{(h_x,h_y)\rightarrow 0} \frac{f(x+h_x, y+h_y)-f(x,y)}{\sqrt{h_x^2+h_y^2}}$

Absolutely not. With this definition the only differentiable functions would be   constant ones; for others this limit fails to exist. The definition of derivative as a limit of difference quotient works only for functions of one variable.
The proper definition of being jointly differentiable at $(x,y)$: there exists a vector $(a,b)$ such that 
$$
\lim_{(h_x,h_y)\rightarrow 0} \frac{|f(x+h_x, y+h_y)-f(x,y)-ah_x-bh_y|}{\sqrt{h_x^2+h_y^2}} =0
$$
This vector $(a,b)$ is the   derivative of $f$ at $(x,y)$. The continuity of derivative means that $a$ and $b$ are continuous functions of $(x,y)$.
It is true that a function has the above property if and only if its  partials exist and they are continuous. But this is a theorem, not a definition.
A: In my experience, this means that each of $f(x,.)$, and $f(.,y)$ is continuously-differentiable. And, since the partials exist and they are continuous, then $f(x,y)$ is itself differentiable. This is the way it is described in the bottom of the first column, in the first page of:  http://www.ams.org/notices/201107/rtx110700896p.pdf
