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In multivariable calculus, given a function like $F(x,y,z) =0$, the implicit function $z=f(x,y)$ exists if and only if $\frac{\partial F}{\partial z} \not = 0$. And the implicit function is given by $\frac{dz}{dy}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$.

The theorem can be proved by mathematical reasoning. But I want to know whether there is some intuitive understanding of the theorem, say geometry intuition in the three dimensional space, etc.

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See the nice (and heuristic) introduction of the book of "The Implicit Function Theorem, History, Theory and Applications" (Birkhauser, 2013) by S. Krantz and H. Parks. The above mentioned linear approach is exposed in detail. – user193056 Nov 16 '14 at 19:33
up vote 5 down vote accepted

First, the IFT is a sufficient condition; if you let $F(x,y) = x-y^3$, then an implicit function exists ($y = \sqrt[3]{x}$), but $\frac{\partial F(x,0)}{\partial y} = 0$.

Regarding intuition, think in terms of linear functions. Just take $F(x,y) = a x + b y$. Notice that $\frac{\partial F(x,y)}{\partial x} = a$, $\frac{\partial F(x,y)}{\partial y} = b$. If $b\neq 0$, then the implicit function is explicitly given by $ y = -\frac{a}{b} x$, from which you can see that $\frac{\partial y(x)}{\partial x} = -\frac{a}{b} = - (\frac{\partial F(x,y)}{\partial y})^{-1} \frac{\partial F(x,y)}{\partial x}$.

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Essentially, the Inverse Function Theorem tells you that if a continuously differentiable function is has an invertible derivative at a point (i.e. the linearization of the function there has an inverse) then the function itself is invertible on some neighborhood of the point. The Implicit Function Theorem is just a special case of the Inverse Function Theorem, which follows from defining $G(x,y,z)=(z,F(x,y,z))$.

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You mean $F(x,y)$ ? – Student Apr 5 '12 at 8:26

If you question ask for the intuition concerning the existence of the implicit function, let me suggest the following.

I prefer to simplify by restricting to two variables. The theorem states that if at $(x_0,y_0)$ the curve $\{(x,y)\in\mathbb R^2 : F(x,y)=0\}$ does not look like a vertical line (i.e if $\partial F/\partial y(x_0,y_0)\neq 0$) then the curve is locally the graph of a function $x\mapsto y(x)$. A good example to have in mind is the circle, $F(x,y)=x^2+y^2-1$. The same reasoning works for higher dimensions.

If you ask about the form of the implicit function, the answer of copper.hat looks nigh enough to me.

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