How to take $\frac{\partial }{\partial x}\langle x,y \rangle$ I confess this is a silly question. I need to take the partial derivatives of $f(x,y):\mathbb{R}^{2n}\to\mathbb{R}$ of $f(x,y) = \langle x,y\rangle$. That is:
$$\frac{\partial }{\partial x}\langle x,y \rangle, \frac{\partial }{\partial y}\langle x,y \rangle$$
But how to view this as a limit? For example:
$$\lim_{t\to 0} \frac{\langle x+t,y\rangle-\langle x, t\rangle}{t}$$
I cannot expand this into anything useful. How do I take this gradient?
 A: For each $v=(v_1,v_2)\in \mathbb{R}^n\times\mathbb{R}^n=\mathbb{R}^{2n}$ we get
\begin{align}
\partial_vf(x,y)&=\lim_{t\to0}\frac{f((x,y)+t(v_1,v_2))-f(x,y)}{t}
=\lim_{t\to0}\frac{1}{t}\left(\langle x+tv_1,y+tv_2\rangle-\langle x,y\rangle\right)\\
&=\lim_{t\to0}\frac{1}{t}\left(t\langle x,v_2\rangle+t\langle v_1,y\rangle+t^2\langle v_1,v_2\rangle\right)=\langle x,v_2\rangle+\langle v_1,y\rangle.
\end{align}
Since $\partial_{(e_i,0)}f(x,y)=x_i$ and $\partial_{(0,e_i)}f(x,y)=y_i$ for $i=1,\ldots,n$ you get
$$
\frac{\partial}{\partial x}f(x,y)=\left(\partial_{(e_1,0)}f(x,y),\ldots,\partial_{(e_n,0)}f(x,y)\right)=x^T
$$
and 
$$
\frac{\partial}{\partial x}f(x,y)=\left(\partial_{(0,e_1)}f(x,y),\ldots,\partial_{(0,e_n)}f(x,y)\right)=y^T
$$
A: Those are not partial derivatives in the usual sense. The usual partial derivatives should be w.r.t one variable only, but in this case, it is w.r.t. $n$ variables. But we can still interpret the symbols in some other ways.
Given a fixed $y\in\Bbb R^n$, define $f_y:\Bbb R^n\to\Bbb R$ by $f_y(x):=\langle x,y \rangle$. Then $\frac{\partial }{\partial x}\langle x,y \rangle$ is interpreted as total derivative of $f_y$. To write it as limit, it would be to find the linear transformation $L:\Bbb R^n\to\Bbb R$ such that $$\lim\limits_{t\to 0}\frac{\lVert\langle x+t,y\rangle-\langle x,y\rangle -Lt\rVert}{\lVert t\rVert}=0$$ where $t\in\Bbb R^n\setminus\{0\}$.
Your interpretation $$\lim_{t\to 0} \frac{\langle x+t,y\rangle-\langle x, t\rangle}{t}$$ is incorrect in several ways. The variable $t$ is either a scalar in $\Bbb R$ or a vector in $\Bbb R^n$, but either interpretation is not possible because you try to add $t$ with $x$ and divide by $t$ simultaneously. Moreover, to find partial derivatives w.r.t. $x$, you should fix the variable $y$, that is you want to calculate the difference $\langle x+t,y\rangle-\langle x, y\rangle$ but not $\langle x+t,y\rangle-\langle x, t\rangle$.
