"Inverse" Chain Rule Green's Theorem, as we know, takes $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$. However, my functions $P$ and $Q$ aren't easily derived in $x$ and $y$ (actually I'm lazy) but are easily so in $r$ and $\theta$. How can I write those partials in terms of the easier ones? I looked online but didn't find much.
It's like an "inverse" Chain Rule, since normally the Chain Rule is used to find $\frac{\partial P}{\partial r}$ and $\frac{\partial P}{\partial \theta}$ in terms of $\frac{\partial P}{\partial x}$ and $\frac{\partial P}{\partial y}$, but in this case I want the opposite.

MY WORK: I gave it a shot and arrived at the following (note, $t$ is $\theta$):
$$\frac{\partial Q}{\partial x}
= \frac{\frac{\partial Q}{\partial r}\frac{\partial y}{\partial t}-\frac{\partial Q}{\partial t}\frac{\partial y}{\partial r}}
{\frac{\partial x}{\partial r}\frac{\partial y}{\partial t}-\frac{\partial x}{\partial t}\frac{\partial y}{\partial r}}$$
This works for me, but is this correct? Can it be simplified further? Also, is this something common?

EDIT: the equation above has been confirmed by @CameronBuie's answer; I applied the same thinking for the 3-dimensional case (that is, a scalar field $u(x,y,z)$ with the coordinate change $(x,y,z)=\mathbf r(r,s,t)$ and arrived at this -- I don't intend this to be useful (it's unwieldy) or correct (didn't check, but seems right), but I thought I might store it here:
$$\frac{\partial u}{\partial z} =
\frac{
\left(
\frac{\partial x}{\partial t}\frac{\partial u}{\partial t}
-
\frac{\partial x}{\partial s}\frac{\partial u}{\partial r}
\right)
\left(
\frac{\partial x}{\partial r}\frac{\partial y}{\partial s}
-
\frac{\partial x}{\partial s}\frac{\partial y}{\partial r}
\right)
-
\left(
\frac{\partial x}{\partial r}\frac{\partial u}{\partial s}
-
\frac{\partial x}{\partial s}\frac{\partial u}{\partial r}
\right)
\left(
\frac{\partial x}{\partial t}\frac{\partial y}{\partial t}
-
\frac{\partial x}{\partial s}\frac{\partial y}{\partial r}
\right)
}{
\left(
\frac{\partial x}{\partial t}
\frac{\partial z}{\partial t}
-
\frac{\partial x}{\partial s}
\frac{\partial z}{\partial r}
\right)
\left(
\frac{\partial x}{\partial r}
\frac{\partial y}{\partial s}
-
\frac{\partial x}{\partial s}
\frac{\partial y}{\partial r}
\right)
-
\left(
\frac{\partial x}{\partial r}
\frac{\partial z}{\partial s}
-
\frac{\partial x}{\partial s}
\frac{\partial z}{\partial r}
\right)
\left(
\frac{\partial x}{\partial t}
\frac{\partial y}{\partial t}
-
\frac{\partial x}{\partial s}
\frac{\partial y}{\partial r}
\right)
}
$$
$\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ can now be found solving the 2x2 system below (chain rules), or by finding a pattern on the formula above.
$$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r}+\frac{\partial u}{\partial z}\frac{\partial z}{\partial r}$$
$$\frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial u}{\partial z}\frac{\partial z}{\partial s}$$
 A: Ultimately, the usual chain rule will do the trick, here. For example if you can express $Q=Q(r,\theta)$, then $$\frac{\partial Q}{\partial x}=\frac{\partial Q}{\partial r}\cdot\frac{\partial r}{\partial x}+\frac{\partial Q}{\partial\theta}\cdot\frac{\partial\theta}{\partial x}.$$

Let's examine your proposed identity: $$\frac{\partial Q}{\partial x}
= \frac{\frac{\partial Q}{\partial r}\frac{\partial y}{\partial\theta}-\frac{\partial Q}{\partial\theta}\frac{\partial y}{\partial r}}
{\frac{\partial x}{\partial r}\frac{\partial y}{\partial\theta}-\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}}\tag{$\star$}$$
I am assuming that $r$ and $\theta$ are expressible as functions of $x$ and $y,$ and that $x$ and $y$ are expressible as functions of $r$ and $\theta.$ I am also assuming that $Q$ is expressible as a function of $x$ and $y$ (alternately, of $r$ and $\theta$) only.
We begin with the clearly true equation $$\frac{\partial Q}{\partial y}\frac{\partial y}{\partial\theta}\frac{\partial y}{\partial r}=\frac{\partial Q}{\partial y}\frac{\partial y}{\partial r}\frac{\partial y}{\partial\theta}.$$ Adding $\frac{\partial Q}{\partial x}\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}$ to both sides, then factoring the left-hand side, we have $$\left(\frac{\partial Q}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial Q}{\partial y}\frac{\partial y}{\partial\theta}\right)\frac{\partial y}{\partial r}=\frac{\partial Q}{\partial x}\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}+\frac{\partial Q}{\partial y}\frac{\partial y}{\partial r}\frac{\partial y}{\partial\theta}.$$ Applying the chain rule to the parenthetical factor on the left-hand side gives us $$\frac{\partial Q}{\partial\theta}\frac{\partial y}{\partial r}=\frac{\partial Q}{\partial x}\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}+\frac{\partial Q}{\partial y}\frac{\partial y}{\partial r}\frac{\partial y}{\partial\theta}.$$ Next, we add $\frac{\partial Q}{\partial x}\frac{\partial x}{\partial r}\frac{\partial y}{\partial\theta}$ to both sides, then group the terms on the right-hand side with a factor of $\frac{\partial y}{\partial\theta},$ yielding $$\frac{\partial Q}{\partial\theta}\frac{\partial y}{\partial r}+\frac{\partial Q}{\partial x}\frac{\partial x}{\partial r}\frac{\partial y}{\partial\theta}=\frac{\partial Q}{\partial x}\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}+\left(\frac{\partial Q}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial Q}{\partial y}\frac{\partial y}{\partial r}\right)\frac{\partial y}{\partial\theta}.$$ Another application of the chain rule to the parenthetical factor on the right-hand side gives us $$\frac{\partial Q}{\partial\theta}\frac{\partial y}{\partial r}+\frac{\partial Q}{\partial x}\frac{\partial x}{\partial r}\frac{\partial y}{\partial\theta}=\frac{\partial Q}{\partial x}\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}+\frac{\partial Q}{\partial r}\frac{\partial y}{\partial\theta},$$ whence a bit more rearrangement and factoring gives us $$\left(\frac{\partial x}{\partial r}\frac{\partial y}{\partial\theta}-\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}\right)\frac{\partial Q}{\partial x}=\frac{\partial Q}{\partial r}\frac{\partial y}{\partial\theta}-\frac{\partial Q}{\partial\theta}\frac{\partial y}{\partial r}.$$
So far, so good. However, at this point, there is an issue that must be addressed: how can we be sure that $\frac{\partial x}{\partial r}\frac{\partial y}{\partial\theta}-\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}$ is non-zero, so that we can divide by it to obtain $(\star)$? Depending on the various functions involved, this may be no small feat!
On the other hand, if $x,y,r,\theta$ are related in the usual way, then computing the partial derivatives of $r$ and $\theta$ with respect to $x$ (or $y$) shouldn't be too difficult. For reference: $$\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}\\\frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}\\\frac{\partial\theta}{\partial x}=-\frac{y}{x^2+y^2}\\\frac{\partial\theta}{\partial x}=\frac{x}{x^2+y^2}$$ At that point, we need only apply relevant substitutions of $r$ and $\theta$ in terms of $x,y,$ then simplify as far as we are able.
The simplicity of each approach will depend largely on the relationships between the variables, and on the functions $P$ and $Q.$ Without knowing more about your particular problem, I'm afraid I can't say more.
