# Implicit Function Theorem and Theorem of Lagrange Multipliers

I don't get how they can take the derivative of g_1 and g_2 with respect to x_1 and x_2 when they are defined as

h_1=h_1 (x_3,x_4,...,x_n )=x_1 and h_2=h_2 (x_3,x_4,...,x_n )=x_2

I need a mathematical justification for how this can be written simply as (21) when x_1 and x_2 are defined as functions from other variables: x_3,x_4,...,x_n

Is it by using the chain rule or something else? Please go through the detailed steps for how this can be a normal quadratic matrix as defined in the implicit function theorem and by that showing why this is an applicable form of the implicit function theorem nonsingular matrix in (21):

© BrainMass Inc. brainmass.com December 20, 2018, 12:25 pm ad1c9bdddfhttps://brainmass.com/math/functional-analysis/implicit-function-theorem-theorem-lagrange-multipliers-596801

#### Solution Preview

Please look at the attached pdf file for a better formatted answer.

{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}

...

#### Solution Summary

We explain and give explicit examples that use the Implicit Function Theorem. Later we explain the use of the Implicit Function Theorem in the proof of the Theorem of Lagrange multipliers.