Implicit Function Theorem and Theorem of Lagrange Multipliers

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{Introduction}
Let us suppose that we have a function F: R^2to R and consider an equation of the form:
F(x,y)=0.
It is natural to ask the question: is it possible to write y as a function of x?

Let us suppose that it is possible, that is, that there exists a function y(x) that satisfies
F(x,y(x))=0
and let us further suppose that F and y(x) are differentiable. Using the chain rule we can differentiate the previous
equation with respect to x:
pd{F}{x}+pd{F}{y}y'(x)=0
from where we obtain
{equation}
label{fyneq0}
y'(x)=-frac{pd{F}{x}}{pd{F}{y}}.
{equation}
Note that in order to make the last calculation it is necessary that pd{F}{y}neq 0.

{example}
Consider the equation
{equation}
label{circle}
x^2+y^2=1.
{equation}
Is it possible to write y as a function of x? The answer is that it is {not} possible to
{globally} write y as a function of x or x as a function of y (this is because the curve described
by eqref{circle} is not the graph of a function as it fails the so called vertical test); but if (x_0,y_0) is a point that
satisfies eqref{circle} and y_0neq 0, then it is possible to write y as a function of x in a neighborhood of (x_0,y_0).
Moreover, differentiating the equation with respect to x as we did before in eqref{fyneq0} we have
{equation}
label{derivative-circle}
y'=-frac{x}{y}.
{equation}

If y_0=0, then it is not possible to write y as a function of x in a neighborhood around the point (x_0,y_0). In this
case we can write x as a function of y instead.

For instance, if we are given the point (1/2,sqrt{3}/2), then y(x)=sqrt{1-x^2} satisfies eqref{circle} for xin[-1,1] and
y(1/2)=sqrt{3}/2. If we are given the point (1/2,-sqrt{3}/2), then y(x)=-sqrt{1-x^2} satisfies eqref{circle} for
xin[-1,1] and y(1/2)=-sqrt{3}/2. So the form of y(x) depends on the point around which we want to write y as a function
of x.

Using eqref{derivative-circle} we can calculate y'(1/2) for (x_0,y_0)=(1/2,sqrt{3}/2):
y'(1/2)=-frac{1/2}{y(1/2)}=-frac{1/2}{sqrt{3}/2}=-frac{1}{sqrt{3}}.
Of course, in this case we can take the explicit expression y=sqrt{1-x^2}, differentiate it, y'=-x/sqrt{1-x^2}, and
evaluate it y'(1/2)=-1/sqrt{3}. The point here is that we can know the derivative at (x_0,y_0) without the need for an
explicit formula for y(x).
{example}
{example}
label{example-system}
Let us consider a more general case, a system of equations. Let as suppose we have variables x_1,...,x_n and
y_1,...,y_m related by m equations
{equation}
label{gis}
g_i(x_1,...,x_n,y_1,...,y_m)=0{ for }1< i< m.
{equation}
We can write such a system in the following way. Define F: R^{n+m}to R^m by
F(x_1,...,x_n,y_1,...,y_m)={pmatrix}
...