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I really want to understand this theorem since it's constantly used in economics for simple equations, which I run into often.

Essentially, we have $\frac{\partial F(x,y)}{\partial x} = z$. What conditions do we need for $x = G(y,z)$ for some function $G$, and what are the partial derivatives of $G$ with respect to $y$ and $z$, as functions of $G$ and $F$?

I understand calculus, but not so much about multidimensional jacobian matrices and such, which is what the wikipedia article on implicit function theorem is centered on.

Jaood
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  • Were you not given a statement of the IFT? It's probably best to work with the one you know. – Git Gud Oct 14 '16 at 15:39
  • No, it's a very informal type book that just keeps referring to random theorems hoping we know them. And most of them we do know (mean value, integration techniques, and so on), but not the implicit function theorem. – Jaood Oct 14 '16 at 15:52
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    Here and here I stated two different versions of the IFT. One more formal than the other, but both more on the formal side of things. Here you can find a few of my answers pertaining to the IFT. You'll find some examples there. Hope this helps. – Git Gud Oct 14 '16 at 16:02

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Suppose that $\frac{\partial F (x_0,y_0)}{\partial x} = z_0$ and that you want to solve $\frac{\partial F (x,y)}{\partial x} = z$ for $y$ and $z$ in a neighborhood of $y_0$ and $z_0$, respectively. This you may do if the linearized equation is solvable and the partial derivative is reasonably smooth. In particular, the slope (if in one variable) $$a = \frac{\partial}{\partial x} \frac{\partial F (x,y)}{\partial x} (x_0,y_0) \neq 0$$ (or in higher dimension should be an invertible matrix). In that case you may solve and obtain as solution $x=G(y,z)$ in some (small) neighborhood of $(y_0,z_0)$. For the derivatives you may use that $$ \frac{\partial F }{\partial x}(G(y,z),y) - z = 0$$ and take derivatives with respect to $y$ and $z$. For example taking a $z$-derivative: $$ \frac{\partial^2 F }{\partial x^2}(G(y,z),y)\frac{\partial G(y,z)}{\partial z} - 1 = 0$$

H. H. Rugh
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