Derivative of a map $ f:\mathbb{R^n\times R^n}\rightarrow\mathbb{R}$ 
I want to calculate the derivative of a function $$f:\mathbb{R^n×R^n}\rightarrow\mathbb{R}$$ defined by
  $$f(x,y)=\langle Ax,y \rangle,$$where $A$ is any $n\times n$ matrix over set of reals $\mathbb{R}$. 

I have never seen such types questions to calculate derivative in which domain is $\mathbb{R^n\times R^n}$. Basically my question is that, 

is the derivative of $f$ is same  as that of 
  function $$g:\mathbb{R^{2n}}\rightarrow\mathbb{R}$$ defined by $$g(x_1,x_2,.......x_n,y_1,y_2,.....y_n)=\langle Ax,y \rangle,$$where $x=(x_1,x_2......x_n)$,$y=(y_1,y_2.....y_n)$.

Your help would be precious to me, thanks in advance!
 A: Don't be overly concerned about the "type" of a point in $\mathbb R^n \times \mathbb R^n$. 
Mathematicians commonly apply the set theoretic "identity" $X^m \times X^n = X^{m+n}$ even if it is not strictly speaking true from a rigorous set theoretic perspective.
Under the covers, one is taking advantage of a silent agreement amongst the mathematical community to employ the bijection 
$$((x_1,...,x_m),(y_1,...,y_n)) \leftrightarrow (x_1,...,x_m,y_1,...,y_n)
$$
A: $$\langle Ax, y\rangle =\sum_{i,j} a_{i,j}x_i y_j$$
Then $\frac{\partial f}{\partial x_\ell} = \sum_j a_{\ell, j} y_j$ and $\frac{\partial f}{\partial y_\ell} = \sum_i a_{i,\ell} x_i$. From this you can construct your Jacobian matrix, which will really here just be a row vector.
A: It depends on whether you want to identify $\mathbb R^n\times\mathbb R^n$ with $\mathbb R^{2n}$ or not. Often there is a reason for the distinction; because of that, below is a computation of the derivative.
At each point $(x,y)$, the derivative is a linear map $D_{x,y}:\mathbb R^n\times \mathbb R^n\to\mathbb R$, such that 
$$\tag1
\lim_{(h,k)\to(0,0)}\frac{f(x+h,y+k)-f(x,y)-D_{x,y}(h,k)}{\|(h,k)\|}=0.
$$ 
In this case, we need
$$\tag2
\lim_{(h,k)\to(0,0)}\frac{\langle A(x+h),y+k\rangle-\langle Ax,y\rangle-D_{x,y}(h,k)}{\sqrt{\|h\|^2+\|k\|^2}}=0,
$$ 
which further becomes 
$$\tag3
\lim_{(h,k)\to(0,0)}\frac{\langle Ah,y\rangle+\langle Ax,k\rangle+\langle Ah,k\rangle-D_{x,y}(h,k)}{\sqrt{\|h\|^2+\|k\|^2}}=0.
$$ 
We have 
$$
\frac{\langle Ah,k\rangle}{\sqrt{\|h\|^2+\|k\|^2}}\leq\frac{\|A\|\,\|h\|\,\|k\|}{\sqrt{\|h\|^2+\|k\|^2}}\leq\|A\|\,\sqrt{\|h\|\,\|k\|}\to0,
$$
so $(3)$ is satisfied if we take $$D_{x,y}(h,k)=\langle Ah,y\rangle+\langle Ax,k\rangle.$$
