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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?

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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 $\frac{\partial x}{\partial x^\prime}=1$

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Yes. There's also a $\frac{\partial A}{\partial y} \frac{\partial y}{\partial x}$, but since that's zero, it's omitted.

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  • $\begingroup$ Thanks. I guess I'm more looking for a pointer to the theory for why this is the case (preferably without getting into jacobians or tensors!) so that I can convince myself. $\endgroup$ – hertzsprung Jul 5 '14 at 22:02
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    $\begingroup$ @hertzsprung: It's all about the (multivariable) chain rule. mathworld.wolfram.com/ChainRule.html $\endgroup$ – Eric Towers Jul 5 '14 at 22:08
  • $\begingroup$ The wolfram link helps, but it looks like the subscripts of $u$ are not consistent... is it $u_i$ or $u_j$? Also, how do $m$ and $n$ correspond to $i$ and $j$, and why do we need two pairs of letters? $\endgroup$ – hertzsprung Jul 17 '14 at 7:51
  • $\begingroup$ @hertzsprung: $i$ and $j$ are placeholders -- two uses of $i$ need not be related (although it is good didactic practice to make them related). At the MathWorld article, the subscript $i$ only appears on $u$ and therefore is ranging over $[1,p]$. The subscript $j$ only appears on $x$ and therefore ranges over $[1,n]$. We need these letters because the chain rule applies to one or more (outermost) dependent variables (the $\{y_i \mid i \in [1,m]\}$) which depend on dependent variables (the $\{u_i \mid i \in [1,p]\}$) which depend on independent variables (the $\{x_i \mid i \in [1,n]\}$). $\endgroup$ – Eric Towers Jul 17 '14 at 15:00

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