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$G_1(x, y_1(x), y_2(x)) = 0$

$G_2(x, y_1(x), y_2(x)) = 0$

Since $G_1$ and $G_2$ are composite functions I can use the chain rule and split them up into partial derivatives. The derivative of both functions should be $0$ since they are both horizontal lines. I'm not sure what to do after that. Maybe just substitution?

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Are you told they are horizontal lines or asserting it? There is nothing here to support that. The derivative with respect to what? Let $G=y-x^2$ We don't have any derivatives that are identically zero. –  Ross Millikan Mar 13 '13 at 4:30
    
You are right. It shouldn't have to be horizontal. It was just a shot in the dark. –  orbis Mar 13 '13 at 4:36
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up vote 1 down vote accepted

Assuming $G,y_1,y_2$ are nice enough to satisfy hypotheses of the implicit function theorem, $$ \partial_xG_1 = \partial_1 G_1 + \partial_2G_1 \partial_xy_1 + \partial_3 G_1 \partial_x y_2 = 0 $$ and $$ \partial_xG_2 = \partial_1 G_2 + \partial_2G_2 \partial_xy_1 + \partial_3 G_2 \partial_x y_2 =0. $$ This is a two-by-two linear system for $\partial_x y_1$ and $\partial_x y_2$ which (I put in Mathematica) solved $$ \partial_xy_1 = -\frac{\partial_1G_2 \partial_3 G_1 - \partial_1 G_1 \partial_3 G_2}{\partial_2G_2 \partial_3 G_1 - \partial_2 G_1 \partial_3 G_2}. $$

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That looks right. I wish I knew a way to solve the system. It gets pretty bad algebraically. Does anyone know any other way? –  orbis Mar 13 '13 at 4:47
    
Don't get lost in the notation. You have something of the form $ax + by = z_1$ and $cx + dy = z_2$. The solution is two lines of linear algebra which I can't be bothered to type. –  nullUser Mar 13 '13 at 4:49
    
I understand. Thanks a bunch. –  orbis Mar 13 '13 at 4:50
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