Given a function of Cartesian coordinates $(x, y)$ in $g(x, y) = \frac{d}{dx}f(x, y)-y\frac{d}{dy}f(x, y)$, I want to transform this expression to polar coordinates $(r,\phi)$. I saw a solution using built-in chain rule here: Change variables in differential expressions and a possible solution using replacement here: Transformation of Derivatives under change of coordinates. I also came across TransformedField , which seems very succinct but it would only tranform $(x, y)$ in the expression but not the derivatives, i.e. it gives me the following:

TransformedField["Cartesian" -> "Polar", 
  D[f[x, y], x] - x*D[f[x, y], y], {x, y} -> {r, \[Phi]}


$$f^{(1,0)}(r \cos (\phi ),r \sin (\phi ))-r \cos (\phi ) f^{(0,1)}(r \cos (\phi ),r \sin (\phi ))$$.

I wonder why this is the case? Or if there is any trick to make it work with TransformedField?

Edit 1

Kuba pointed out an existing function in "MoreCalculus," which works well for simple expression. But when I use it for a more complicated expression, I'm not sure if it outputs correct result:

Input expression with $dx = \frac{d}{dx}$ and $dy = \frac{d}{dy}$ Input

Output is very long but contains an expression with prime as shown here: $$i e \left(-\frac{1}{2} B r \sin (\phi )\right)'$$

And I'm not sure what that prime denotes or why it doesn't simplify (I already used FullSimplify). Any help would be appreciated.

  • $\begingroup$ DChange[ D[f[x, y], x] - x*D[f[x, y], y], "Cartesian" -> "Polar", {x, y}, {r, \[Phi]}, f[x, y] ] seems to do the trick. Is that the answer: change of variables in differential expressions? $\endgroup$ – Kuba Jan 11 at 23:11
  • $\begingroup$ Thanks for the link Kuba! So I applied it to my more complicated expression , and one of the outputs has a prime on it (shown in the edits). Do you know what the prime means? $\endgroup$ – Histoscienology Jan 12 at 2:53
  • $\begingroup$ Please provide a copyable code for input and output. After I created that function I used it maybe twice, and it has around four unit tests so I am not surprised fails for more complicated cases ;) $\endgroup$ – Kuba Jan 12 at 20:38

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