Following lecture notes from MIT it says that, given some variable $A = A(x, y, z(x, y, r, t), t)$ where $r$ is a transformed vertical coordinate
$\left. \frac{\partial A}{\partial x} \right|_r = \left. \frac{\partial A}{\partial x} \right|_z + \frac{\partial A}{\partial z} \left. \frac{\partial z}{\partial x} \right|_r $
I can see this works by trying concrete examples, and I think I can see the second term on the right is due to the chain rule (because $A$ is a function of $z$ is a function of $x$). But where does the first term on the right come from? Is it because $A$ is also a function of $x$ explicitly? Can someone point me to a piece of theory that underpins this?
I think that the confusion comes from transforming from $(x,z) \rightarrow (x,r)$. These 2 $x$s are not the same in terms of partial derivatives because, the first assumes that $z$ remains constant and the second assumes that $r$ remains constant. So if instead we were to start with: $$A(x(x^\prime), z(x^\prime,r))$$ Then the chain rule would give us: $$\frac{\partial A}{\partial x^\prime} =\frac{\partial A}{\partial x} \frac{\partial x}{\partial x^\prime} +\frac{\partial A}{\partial z} \frac{\partial z}{\partial x^\prime}$$ where what you previously wrote as $\frac{\partial A}{\partial x}\big\bracevert_r$ is now written $\frac{\partial A}{\partial x^\prime}$ and what you previously wrote as $\frac{\partial A}{\partial x}\big\bracevert_z$ is now written as just $\frac{\partial A}{\partial x}$. And of course $\frac{\partial x}{\partial x^\prime}=1$