Where's my error in this partial derivatives problem? Let $u(x, y)=x+y$. What is $\displaystyle\frac{\partial u}{\partial x}$ and $\displaystyle\frac{\partial u}{\partial y}$? My answers are $1$ and $1$.
Suppose I now told you that $y=x$, so that $u=2x$.
Now it appears that $\displaystyle\frac{\partial u}{\partial x}=2$ and $\displaystyle\frac{\partial u}{\partial y}=0$.
Where have I gone wrong?
Note: My question is completely different from "Partial derivatives paradox". I have renamed my question.
 A: When we have two variables function $u(x,y)=x+y$, we discuss about a surface in $\mathbb{R}^3$. The derivation $\displaystyle\frac{\partial u}{\partial x}$ means derivation in $x$'s axis direction, and  $\displaystyle\frac{\partial u}{\partial y}$ is in $y$'s. this make a derivative 
$$\nabla u=\left(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}\right)=\frac{\partial u}{\partial x}\vec{i}+\frac{\partial u}{\partial y}\vec{j}$$
for $u$.
But when you have $y=x$, then there is a one-variable function $u=2x$ with a graph in $\mathbb{R}$. Also the derivation isn't a 2D vector and actually $\displaystyle\frac{d u}{d x}=2$. So we discuss about a differnt function with previous 3D function.
A: Here's a way to think about it that may help.  You start with a function $u:\mathbb{R}^2\longrightarrow\mathbb{R}^1$  When you talk about setting $x=y$ you are now talking about the function $u\circ h$ where $h:\mathbb{R}^1\longrightarrow\mathbb{R}^2$ is the function defined by $h(x)=(x,x)$.  To differentiate $u\circ h,$ you have to use the chain rule.
I'm sure you wouldn't ask, "When I have $f(x)=x^2$ I have $f'(x)=2x,$ but when I set $x=\sin x$ I get ... ."  Same thing.
A: Back in beginning calculus you were used to seeing things like "$u=2x$". There it was understood that when such a thing appeared, it meant that $u$ is a function of the one real variable $x.$ But now $u(x+y)=x+y$ is a function of two real variables. It is true that in this case $u(x,x)=2x,$ but abbreviating that observation by writing $u=2x,$ without further clarification, would alarm even the woolier ranks of engineers, where things are wrong half the time on a good day. So: $d\,u(x,x)/dx = 2x$ is correct, but $du/dx = 2$ is so false it's not even wrong.
