Suppose I know
\begin{align} dX = A dx + B dy \end{align} Then I know \begin{align} \frac{\partial X}{\partial x} =A, \quad \frac{\partial X}{\partial y} =B \end{align} Then is there any smart way to compute \begin{align} \frac{\partial x}{\partial X}, \quad \frac{\partial y}{\partial X} \end{align}
What I know :
Take simple example "two dimensional polar system" \begin{align} &x= r\cos(\theta), \quad y = r\sin(\theta) \end{align} \begin{align} &dx = \cos(\theta) dr - r \sin(\theta) d\theta \\ &dy = \sin(\theta) dr + r\cos(\theta) d\theta \end{align} in terms of $dr, d\theta$, by multiplying inverse metric for coefficients, I have \begin{align} &dx = \cos(\theta) dr - r \sin(\theta) d\theta \\ &dy = \sin(\theta) dr + r\cos(\theta) d\theta \end{align} Then Finally \begin{align} &\frac{\partial x}{\partial r} = \frac{x}{r}=\cos(\theta), \quad \frac{\partial r}{\partial x} =\cos(\theta) \\ &\frac{\partial y}{\partial r} =\frac{y}{r} =\sin(\theta), \quad \frac{\partial r}{\partial y} =\sin(\theta) \\ & \frac{\partial x}{\partial \theta} =-y = -r\sin(\theta), \quad \frac{\partial \theta}{\partial x} =-\frac{1}{r} \sin(\theta) \\ & \frac{\partial y}{\partial \theta} = x = r\cos(\theta), \quad \frac{\partial \theta}{\partial y} = \frac{\cos(\theta)}{r} \end{align} It seems they related somehow but how?
$\def\o{\tt1}\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}\def\p#1#2{\frac{\partial #1}{\partial #2}}$The problem is that single equation does not provide enough information. The desired calculation requires the full system of equations: $$\eqalign{ dX &= A\,dx + B\,dy \\ dY &= C\,dx + D\,dy \\ }$$ Now calculate the inverse of the coefficient matrix $$\eqalign{ M &= \m{A&B\\C&D} \quad\implies M^{-1} &= \m{E&F\\G&H} \implies \m{dx = E\,dX + F\,dY \\dy = G\,dX + H\,dY} \\ }$$ and extract the desired gradient $$\eqalign{ \p{x}{X} &= E \\ }$$ These block matrix inversion formulas can be used to express $E$ in terms of $(A,B,C,D)$ by treating them as $(\o\times\o)$ matrices.