Change of variables and the partial derivative From time to time, I suddenly get confused with a change of variables in a partial derivative.
Here, I am trying to perform a change of variables $(x,t) \mapsto (\xi, \eta)$ where
$$\xi = t \qquad \qquad \text{and} \qquad \qquad \eta = x+t$$
The question is, how to compute $$\frac{\partial u}{\partial t}$$ in the new coordinate system?
Intuitively, since $\xi = t$ (or rather $t=\xi$), we should have
$$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \xi}$$
However, applying the chain rule for partial derivatives, we instead get
$$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial t} = \frac{\partial u}{\partial \xi}(1) + \frac{\partial u}{\partial \eta}(1) = \frac{\partial u}{\partial \xi} + \frac{\partial u}{\partial \eta}$$
So which one is correct?
 A: The second one is correct. Why is the other one wrong? Well when you write $$\frac{\partial u}{\partial \xi}$$what you really are saying is $$\frac{\partial u}{\partial \xi}{\huge|}_\eta$$, i.e. taking the derivative with respect to $\xi$, but keeping $\eta$ fixed. If we do this, we are restricting ourselves to the path $x=-t$, where $t$ varies. But we wanted to represent $$\frac{\partial u}{\partial t}{\huge|}_x$$with a fixed $x$. So this expression fails to express the same quantity. Using the chain rule works to ensure this error does not happen.
A: What follows is not only the correct way to handle this issue (most of the time people resort to what are to me confusing notational abuses and artifacts from a time when it was hard to be a bit more rigorous), it's a sketch of the answer that I'd like to be given.
To simplify, in what's below I will only consider the whole spaces as domains, but it can be generalized to any open subset of these spaces.

Theorem. Let $m, n$ and $p$ be natural numbers, and $F\colon \mathbb R^n\to \mathbb R^m$ and $G\colon\mathbb R^m\to \mathbb R^p$ differentiable functions. If $H=G\circ F$, then $H$ is differentiable and for all $X$ in $\mathbb R^n$ it holds that $$(DH)_X=(DG)_{F(X)}(DF)_X \tag 1$$

Considering that $F, G$ and $H$ can be written as
$$
F=(f_1, \ldots, f_m),\\
G=(g_1, \ldots, g_p),\\
H=(h_1, \ldots , h_p)
$$
where:


*

*$f_1, \ldots, f_m$ are real functions whose domain is $\mathbb R^n$ and whose partial derivatives all exist,

*$g_1, \ldots, g_p$ are real functions whose domain is $\mathbb R^m$ and whose partial derivatives all exist,

*$h_1, \ldots, h_p$ are real functions whose domain is $\mathbb R^n$ and whose partial derivatives all exist,


$(1)$ can be rewritten as

$$
\begin{bmatrix}
\partial _1h_1 & \ldots & \partial_nh_1\\
\vdots & \ddots & \vdots\\
\partial_1h_p & \ldots & \partial_nh_p
\end{bmatrix}_X
=
\begin{bmatrix}
\partial _1g_1 & \ldots & \partial_mg_1\\
\vdots & \ddots & \vdots\\
\partial_1g_p & \ldots & \partial_mg_p
\end{bmatrix}_{F(X)}
\begin{bmatrix}
\partial _1f_1 & \ldots & \partial_nf_1\\
\vdots & \ddots & \vdots\\
\partial_1f_m & \ldots & \partial_nf_m
\end{bmatrix}_X
\tag{2}
$$


Armed with this formulation of this standard theorem, all that's left is to look at the problem through this lens.
As I understand it you have a differentiable function $u\colon \mathbb R^2\to \mathbb R$, a function $\varphi \colon \mathbb R^2\to\mathbb R^2, (x,t)\mapsto (t, x+t)$ (which is also differentiable) and you're asked to find $\partial _2(u\circ \varphi)$.
Now it's easy. In the notation above we have $m=2=n$, $p=1$, $G=u$, $F=\varphi$, $f_1\colon \mathbb R^2\to \mathbb R, (x,t)\mapsto t$ and $f_2\colon \mathbb R^2\to \mathbb R, (x,t)\mapsto x+t$. Set $H=u\circ \varphi$ and conclude that for all $(x,t)$ in $\mathbb R^2$ it holds that
$$
\begin{align}
\begin{bmatrix}
\partial _1h_1 &  \partial_2h_1
\end{bmatrix}_{(x,t)}
&=
\begin{bmatrix}
\partial _1u & \partial_2u
\end{bmatrix}_{F(x,t)}
\begin{bmatrix}
\partial _1f_1 & \partial_2f_1\\
\partial_1f_2 & \partial_2f_2
\end{bmatrix}_{(x,t)}\\
&=
\begin{bmatrix}
\left(\partial _1u\right)(t,x+t) & \left(\partial_2u\right)(t,x+t)
\end{bmatrix}
\begin{bmatrix}
0& 1\\
1 & 1
\end{bmatrix}\\
&=
\begin{bmatrix}
\left(\partial _2u\right)(t,x+t) & \left(\partial _1u\right)(t,x+t)+\left(\partial_2u\right)(t,x+t)
\end{bmatrix}_.
\end{align}
$$
A: Consider $u(x,t)=x^2+t^2$ and $\xi=t, \eta=x+t$. 
Then: $u(\xi(x,t),\eta(x,t))=(\eta-\xi)^2+\xi^2$. We get:
$$u_t=2(\eta-\xi)\cdot (\eta_t-\xi_t)+2\xi\cdot \xi_t=\\
2(x+t-t)\cdot (1-1)+2t\cdot 1=\\
2t,$$
which is true:
$$u_t=(x^2+t^2)_t=2t.$$
