To prove that a functions has partial derivatives every partial has to exist, and every partial exist only if the limit of definition of partial exist. Is this right?
Then if partials exist ,and the limit of derivability involving the Jacobian is equal to zero, (see Wikipedia), then you can say it is derivable. Do you have to check this limit every time to see if is differentiable? Or there is another procedure?
And how to check the partial derivatives are continuous? Do you just see if the partials have one or some points of discontinuity by just looking at the partials functions, and then if they are to be continuous they are differentiable? And if is a piecewise function every branch of every partial must be equal for where f changes between branchs? In the following 2 examples I try to address this questions.
If f is define as$$f(x,y) =\begin{cases} x^2+2x+5y+10 & \text{ for } (x,y)>= (0,0) \\ y^2+2y+x+10 & \text{ for } (x,y)<(0,0) \end{cases}$$
And its partials $$f^\prime_x(x,y) =\begin{cases} 2x+2 & \text{ for } (x,y)>=(0,0) \\ 1& \text{ for } (x,y)<(0,0) \end{cases}$$
$$f^\prime_y(x,y) =\begin{cases} 5 & \text{ for } (x,y)>= (0,0) \\ 2y+2 & \text{ for } (x,y)<(0,0) \end{cases}$$
Can I do this, do a branch partials like above? And the partials are continuous? I think not because $$f^\prime_x(0,0) =\begin{cases} 2 & \text{ for } (x,y)>=(0,0) \\ 1& \text{ for } (x,y)<(0,0) \end{cases}$$ $$f^\prime_y(0,0) =\begin{cases} 5 & \text{ for } (x,y)>= (0,0) \\ 2 & \text{ for } (x,y)<(0,0) \end{cases}$$ so they are not continuous on point (0,0) is that right?
However f is continuous? Plug in (0,0) in the function and you get $$f(x,y) =\begin{cases} 10 & \text{ for } (x,y)>= (0,0) \\ 10 & \text{ for } (x,y)<(0,0) \end{cases}$$ However, I am not convince for example 1st and 3rd quadrants I can see is true but for 2d and 4th wish function takes place?
Now if f was slightly different $$g(x,y) =\begin{cases} x^2+2x+3y+10 & \text{ for } (x,y)= (0,0) \\ y^2+3y+2x+5 & \text{ for } (x,y)\neq (0,0) \end{cases}$$
Its partials
$$g^\prime_x(x,y) =\begin{cases} 2x+2 & \text{ for } (x,y)=(0,0) \\ 2& \text{ for } (x,y) \neq (0,0) \end{cases}$$
$$g^\prime_y(0,0) =\begin{cases} 3 & \text{ for } (x,y)= (0,0) \\ 2y+3 & \text{ for } (x,y) \neq (0,0) \end{cases}$$
$$g^\prime_x(0,0) =\begin{cases} 2 & \text{ for } (x,y)=(0,0) \\ 2& \text{ for } (x,y) \neq (0,0) \end{cases}$$
$$g^\prime_y(0,0) =\begin{cases} 3 & \text{ for } (x,y)= (0,0) \\ 3 & \text{ for } (x,y) \neq (0,0) \end{cases}$$
So it seems that the partials are continuous therefore g is derivable therefore continuous but g is not continuous on(0,0) $$g(0,0) =\begin{cases} 10 & \text{ for } (x,y)= (0,0) \\ +5 & \text{ for } (x,y)\neq (0,0) \end{cases}$$ so it is not derivable and hence its partials are not continuous but they are.
So what is wrong in this second case and with first one? If you can please provide with a detailed answer to the 2 examples, I would appreciate it.
If a function is discontinuous at some point then is no point talk about partials and hence derivative. A function can have a partials however they could not be define ate some points and hence discontinuous partials .however if at points where the partials could be discontinuos exist like $f^\prime_x =\lim_{h \to 0}$ $\frac{f(h,0)-f(0,0)}{h}$ then is said to be derivable at that point (?however partials could be discontinuous?)
Then if the partials are define at every point is said to have continuous partials and hence derivable.
For a picewise function besides to check continuity in each branch you have to check continuity where piecewise function change between branches and then if is continuous you have see if the partials are continues at those points betweens branches.
For the first case
Since the function is not continuous on (0,0) because the function is not defined at (-x,y) and (x,-y) ( by another words is not define in a full neiberhood at origin) so the partials are not define, but even f was then we check if a function has partials, and it has, however they are not continuous as obvious express in the question
For the 2sd case
First of all, you can just write g(x,y)=10 if (x,y)=(0,0), there is no point in writing that way. This way you can also see that is pointless in differentiating x2+2x+3y+10 with the usual rules, as it is only valid in one point and derivatives deal with neighbourhoods. Finally, the "contradiction" is explained by remembering that the usual rules for differentiating only make sense if the derivative is defined, and if the function is not continuos they are not defined.