Having $y=f(x,p)$ and $x=g(p,y) $ and considering we already know $\frac{\partial f}{\partial x}; \frac{\partial f}{\partial p}; \frac{\partial g}{\partial p}; \frac{\partial g}{\partial y} $
Find $\frac{dy}{dx} $ using the 4 derivatives above
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Are you sure about the functions because now you have $y=f(g(p,y),p) = f(g(p,f(x,p))) = ...$ that would never end I think. – Leviathan Jun 10 '17 at 14:05
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1@Leviathan, if we define $F, G : \mathbb R^3 \to \mathbb R$ by $F(x,y,p) = f(x,p) - y$ and $G(x,y,p) = g(p,y) - x$, we have two equations, $F(x,y,p)=0$ and $G(x,y,p) = 0$, that each define a surface in $\mathbb R^3$. The intersection of these is a curve, and we want to know how $y$ changes along that curve when $x$ changes a little. – md2perpe Jun 10 '17 at 14:10
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Differentiating the equations, we have $$dy = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial p} dp$$ $$dx = \frac{\partial g}{\partial p} dp + \frac{\partial g}{\partial y} dy$$
Now we eliminate $dp$ by first multiplying the equations with $\frac{\partial g}{\partial p}$ and $\frac{\partial f}{\partial p}$ respectively: $$\frac{\partial g}{\partial p} dy = \frac{\partial f}{\partial x} \frac{\partial g}{\partial p} dx + \frac{\partial f}{\partial p} \frac{\partial g}{\partial p} dp$$ $$\frac{\partial f}{\partial p} dx = \frac{\partial f}{\partial p} \frac{\partial g}{\partial p} dp + \frac{\partial f}{\partial p} \frac{\partial g}{\partial y} dy$$ and then subtracting: $$\frac{\partial g}{\partial p} dy - \frac{\partial f}{\partial p} dx = \frac{\partial f}{\partial x} \frac{\partial g}{\partial p} dx - \frac{\partial f}{\partial p} \frac{\partial g}{\partial y} dy.$$
Collecting $dy$ on the left side and $dx$ on the right side, we get $$\left( \frac{\partial g}{\partial p} + \frac{\partial f}{\partial p} \frac{\partial g}{\partial y} \right) dy = \left( \frac{\partial f}{\partial p} + \frac{\partial f}{\partial x} \frac{\partial g}{\partial p} \right) dx$$
Thus, $$\frac{dy}{dx} = \frac{ \frac{\partial f}{\partial p} + \frac{\partial f}{\partial x} \frac{\partial g}{\partial p} }{ \frac{\partial g}{\partial p} + \frac{\partial f}{\partial p} \frac{\partial g}{\partial y} }$$
md2perpe
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Using differentials in this way might be a bit unrigorous, but it's just to think of them as derivatives w.r.t. some unnamed parameter. – md2perpe Jun 11 '17 at 08:22