How to find partial derivatives in very abstract case when F is just F(x1,x2) and how to express it correctly? I've just started learning differential calculus and there's one task that I don't completely understand. It sounds like: 
"Given $Y=F(x_1,x_2)+f(x_1)+g(x_2)$, find $\frac {\partial Y}{\partial X_1}$, $\frac {\partial ^2 Y}{\partial X_1^2}$ and $\frac {\partial ^2 Y}{\partial X_1 \partial X_2}$". 
They don't describe exactly what do the F, f and g functions look like.
Still, it's quite easy to find the partial derivative for the $f(x_1)$ and $g(x_2)$ ($\frac {\partial Y}{\partial X_1} f(x_1) = f'(x_1)$ and $\frac {\partial Y}{\partial X_1} g(x_2) = 0$ as $x_2 = const$. 
But what to do with the "big" two-arguments function? What notation to use? Will
$$\frac {\partial Y}{\partial X_1} F(x,y) = \frac {\partial Y}{\partial X_1} F(x,y)_{x_2}$$ be just the required answer? ($x_2$ subscript means that we consider it a constant in this case)
But how to write the answers for $\frac {\partial ^2 Y}{\partial X_1^2}$ and $\frac {\partial ^2 Y}{\partial X_1 \partial X_2}$? Will the result for $\frac {\partial ^2 Y}{\partial X_1^2}$ look just like the one above? And what for the $\frac {\partial ^2 Y}{\partial X_1 \partial X_2}$?
I've tried WolframAlpha just to give me some clues, but it uses some notation I don't understand:
$$\frac {\partial Y}{\partial X_1} F(x,y) = F^{(1,0)}(x,y)$$
I'm getting quite confident when it comes to differentiating some "concrete" functions but I somehow get stuck when it comes to such abstract situations.
Will appreciate any help,
Thanks,
Paul

$$Y=F(x_1,x_2)+f(x_1)+g(x_2) \\ \implies \frac{\partial}{\partial x_1} Y = \frac{\partial}{\partial x_1} \left[F(x_1,x_2)+f(x_1)+g(x_2)\right] \\ \begin{align}\implies \require{enclose}\enclose{box}{\frac{\partial Y}{\partial x_1}} &= \frac{\partial F(x_1,x_2)}{\partial x_1}+ \frac{\partial f(x_1)}{\partial x_1}+\frac{\partial g(x_2)}{\partial x_1}  \\ &= \frac{\partial F(x_1,x_2)}{\partial x_1}+ \frac{df(x_1)}{dx_1}+0 \\ &\enclose{box}{= \frac{\partial F(x_1,x_2)}{\partial x_1}+ \frac{df(x_1)}{dx_1}}\end{align} $$
