I'm struggling to determine two partial derivatives of the following two implicitly-defined, coupled equations.
Given: $$x = X(x, y, E) = f_x(E) + g_x(x, y)$$ $$y = Y(x, y, E) = f_y(E) + g_y(x, y)$$
Find: $\frac{\partial x}{\partial E}$ and $\frac{\partial y}{\partial E}$ in terms of $f_x$, $f_y$, $g_x$, and $g_y$ derivatives (i.e. $\frac{d f_x}{d E}$, $\frac{d f_y}{d E}$, $\frac{\partial g_x}{\partial x}$ $\frac{\partial g_x}{\partial y}$ $\frac{\partial g_y}{\partial x}$ $\frac{\partial g_y}{\partial y}$).
where E is the independent variable.
I've tried deriving each term with respect to E, but end up with "recursion" when deriving $g$ and I'm not sure where to go with that. For example for $x$: $$ \frac{\partial x}{\partial E} = \frac{d f_x}{d E} + \frac{\partial g_x}{\partial x} \frac{\partial x}{\partial E} + \frac{\partial g_x}{\partial y} \frac{\partial y}{\partial E} $$
The "recursion" of $\frac{\partial x}{\partial E}$ doesn't both me (I can just rearrange and take out a common factor) but I don't know how to eliminate $\frac{\partial y}{\partial E}$
I've figured it out and it's not too bad, just a little messy.
Define:
$$a = \frac{d f_x}{d E}; b = \frac{d f_y}{d E};c = \frac{\partial g_x}{\partial x};d = \frac{\partial g_x}{\partial y};e = \frac{\partial g_y}{\partial x};f = \frac{\partial g_y}{\partial y}$$
and let the unknowns be: $$p = \frac{\partial x}{\partial E};q = \frac{\partial y}{\partial E}$$
we then have two equations: $$p = a + cp + dq$$ $$q = b + ep + fq$$
Rearranging: $$p = \frac{a + dq}{1 - c}$$ $$q=\frac{b+ep}{1-f}$$
Solving using Wolfram Alpha: $$p=\frac{a-\frac{bd}{f-1}}{1-c + \frac{de}{f-1}}$$