What is the actual proof for implicit differientiation? ?

Question

I want theory details, not examples. How is it proven?

Answer 1

I'll just discuss the one-dimensional case, but essentially the same things work in higher dimensions. Anyway, suppose F:R^2 -> R is a continuously differentiable function. Consider the level set of F:

F(x,y) = 0.

Now, if you assume that at least for x in some small open set this level set can be locally parametrized by a differentiable function y = g(x), then:

F(x,g(x)) = 0.

Differentiating both sides and using the chain rule gives

dF/dx(x,g(x)) + dF/dy(x,g(x))* dg/dx(x) = 0.

If dF/dy(x,g(x)) is not zero then you can solve for dg/dx(x):

dg/dx(x) = (-dF/dx(x,g(x)))/(dF/dy(x,g(x))).

This is implicit differentiation. The question that remains however is, under what conditions on F(x,y) can you assume that the level set F(x,y) = 0 is given by (x,g(x)). Said another way, when can you assume that the implicit relation F(x,y) = 0 defines y as a function of x? From the above calculation, one natural guess for such a condition would be that dF/dy(x,g(x)) not equal zero, and in fact this is a sufficient condition to prove that y is a function of x at least locally. This result is called The Implicit Function Theorem and is quite important in a number of areas of mathematics.

http://en.wikipedia.org/wiki/Implicit_fu...