Partial Differential Equation Change of Variables Problem Statement
I have a partial differential equation given by
$$\frac{\partial f}{\partial x}=0\tag{1}\label{1}$$
where $f=f(x,y)$. If we introduce a new set of coordinates $q_1$ and $q_2$ such that
$$
\begin{array}{c}
q_1=q_1(x,y) \\ 
q_2=q_2(x,y) \\ 
\end{array}\tag{2}\label{2}
$$
and $g(q_1,q_2)$ corresponds to $f(x,y)$, how do I convert Eq. $\ref{1}$ which is in terms of $f$, $x$, and $y$ to an equation in terms of $g$, $q_1$, and $q_2$?
Solution Attempt
Taking
$$f(x,y) = f(x(q_1,q_2),y(q_1,q_2))=g(q_1,q_2)\tag{3}\label{3}$$
I can substitute equation $\ref{3}$ into equation $\ref{1}$. I think my problem starts when I try to take $\frac{\partial g}{\partial x}$ because the following seems wrong:
$$ \frac{\partial f}{\partial x}=\frac{\partial g}{\partial x}=\frac{\partial g}{\partial q_1}\frac{\partial q_1}{\partial x}\tag{4}\label{4}$$
I think I'm missing some form of the chain rule and believe $\ref{4}$ should really be something like
$$ \frac{\partial f}{\partial x}=\frac{\partial g}{\partial x}=\frac{\partial g}{\partial q_1}\frac{\partial q_1}{\partial x}+\frac{\partial g}{\partial q_2}\frac{\partial q_2}{\partial x}\tag{5}\label{5}$$
because if I make a small change in x, it may cause small changes in $q_1$ and $q_2$, but I don't know how to prove this or if it is true.
Hopefully, this question here will help me answer this question I asked previously. Any help appreciated and thanks in advance!
 A: Yes, $g$ depends on $q_1$ and $q_2$, which both depend on $x, y.$
The chain rule says:
$$\frac{\partial}{\partial x} g(q_1,q_2) = \frac{\partial g}{\partial q_1}\frac{\partial q_1}{\partial x} + \frac{\partial g}{\partial q_2}\frac{\partial q_2}{\partial x},$$
which is what you need.
A: I think I am trying to find notation that made sense and I think I figured out how to do it using this post.
Start by defining the functions we are using
$$f: \Bbb{R}^2 \to \Bbb{R}$$
$$g: \Bbb{R}^2 \to \Bbb{R}$$
and write Eq. 1 more precisely as
$$(\partial_1f)(x,y) = 0$$
This means we want to take the derivative of $f$ with respect to the first argument, $x$. After substituting $g$ in for $f$ in the above equation, we now wish to find how $g$ changes with changes in $x$. This requires us to use the chain rule as shown below:
$$
\begin{array}{c}
(\partial_1f)(x,y)= [(\partial_1g)(q_1(x,y),q_2(x,y))]\cdot(\partial_1q_1)(x,y) \\ 
+ [(\partial_2g)(q_1(x,y),q_2(x,y))]\cdot(\partial_1q_2)(x,y)\\ 
\end{array}$$
Or using sloppy notation:
$$\frac{\partial f}{\partial x}=\frac{d f}{d x}=\frac{d g}{d x}=\frac{\partial g}{\partial q_1}\frac{d q_1}{d x}+\frac{\partial g}{\partial q_2}\frac{d q_2}{d x}$$
Note how I purposely used $d$ instead of $\partial$ in the second and third terms and did the same for $g$ because in each case, we want to know how each function varies with changes in $x$.
