# Implicit Function Theorem

• Oct 14th 2010, 06:50 PM
lilaziz1
Implicit Function Theorem
Hey everyone I have a question on the proof of Implicit Function Theorem. My book says, "Suppose that z is given implicitly as a function z = f(x,y) by an equation of the form F(x,y,z) = 0. This means that F(x,y,f(x,y)) = 0 for all (x,y) in the domain of f. If F and f are differentiable, then we can use the Chain Rule to differentiate the equation F(x,y,z) = 0 as follows:

$\frac{\partial F}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$

But $\frac{\partial}{\partial x}(x) = 1$ and $\frac{\partial}{\partial x}(y) = 0$

So the equation becomes

$\frac{\partial F}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$

My question is, how does $\frac{\partial}{\partial x}(y) = 0$? It doesn't make sense. Is it just saying that let/assume $\frac{\partial}{\partial x}(y) = 0$? Confuseddd.

• Oct 14th 2010, 07:24 PM
Gusbob
Quote:

Originally Posted by lilaziz1
Hey everyone I have a question on the proof of Implicit Function Theorem. My book says, "Suppose that z is given implicitly as a function z = f(x,y) by an equation of the form F(x,y,z) = 0. This means that F(x,y,f(x,y)) = 0 for all (x,y) in the domain of f. If F and f are differentiable, then we can use the Chain Rule to differentiate the equation F(x,y,z) = 0 as follows:

$\frac{\partial F}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$

But $\frac{\partial}{\partial x}(x) = 1$ and $\frac{\partial}{\partial x}(y) = 0$

So the equation becomes

$\frac{\partial F}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$

My question is, how does $\frac{\partial}{\partial x}(y) = 0$? It doesn't make sense. Is it just saying that let/assume $\frac{\partial}{\partial x}(y) = 0$? Confuseddd.

Hey, look at the conditions set at the beginning. y has no x terms in it (it is not a function of x), so if you differentiate it with respect to x, it is regarded as a constant and goes to 0. A similar does not occur for z because z is a function of x (z = f(x,y))
• Oct 14th 2010, 07:25 PM
tonio
Quote:

Originally Posted by lilaziz1
Hey everyone I have a question on the proof of Implicit Function Theorem. My book says, "Suppose that z is given implicitly as a function z = f(x,y) by an equation of the form F(x,y,z) = 0. This means that F(x,y,f(x,y)) = 0 for all (x,y) in the domain of f. If F and f are differentiable, then we can use the Chain Rule to differentiate the equation F(x,y,z) = 0 as follows:

$\frac{\partial F}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial F}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$

But $\frac{\partial}{\partial x}(x) = 1$ and $\frac{\partial}{\partial x}(y) = 0$

So the equation becomes

$\frac{\partial F}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$

My question is, how does $\frac{\partial}{\partial x}(y) = 0$? It doesn't make sense. Is it just saying that let/assume $\frac{\partial}{\partial x}(y) = 0$? Confuseddd.

It is implicitely assumed (pun intended) that x,y are independient variables, so both $\frac{\partial y}{\partial x}=\frac{\partial x}{\partial y}=0$

Tonio
• Oct 15th 2010, 06:06 AM
lilaziz1
Quote:

Originally Posted by Gusbob
Hey, look at the conditions set at the beginning. y has no x terms in it (it is not a function of x), so if you differentiate it with respect to x, it is regarded as a constant and goes to 0. A similar does not occur for z because z is a function of x (z = f(x,y))

That kind of make sense, but can't you solve F(x,y,f(x,y)) for y?
Let's do an example:

Let's say F(x,y,f(x,y)) = 3x + y + z = 0 or something.
Now if you take $\frac{dx}{dx}$ you would get:

$x = \frac{-y-z}{3}$

$x = 0?$
• Oct 15th 2010, 03:35 PM
Gusbob
Quote:

Originally Posted by lilaziz1
That kind of make sense, but can't you solve F(x,y,f(x,y)) for y?
Let's do an example:

Let's say F(x,y,f(x,y)) = 3x + y + z = 0 or something.
Now if you take $\frac{dx}{dx}$ you would get:

$x = \frac{-y-z}{3}$

$x = 0?$

Sorry I should have made it clearer. I think a main point of confusion here is that we're dealing with partial derivatives (i.e. dy/dx is a partial derivative, so is dz/dx)

What does it mean that z is defined implicitly as a function of x and y?
For your equation, it means z must be a function of x and y that satisfies 3x + y + z = 0. So

z = -(3x + y). That is the explicit equation of your F(x,y,f(x,y)) = 0

Now note that the derivative $\frac{dz}{dx} = -3$

So if we go with what you suggested, that we solve F(x,y,f(x,y)) = 0 for y

y = -3x - z.

differentiating with respect to x gives you

$\frac{dy}{dx} = -3 - \frac{d}{dx}z = -3 - (-3) = 0$

Do you see now why $\frac{dy}{dx}$ will always be zero?
• Oct 15th 2010, 04:38 PM
lilaziz1
:O! Yes it does!

This seems a bit recursive though. We're basically saying:

y = -3x - z, where z = -(3x+y) and by direct substitution, we get y = -3x + (3x+y) = y

The partial derivative of y = y with respect to x is 0.

In that case, wouldn't dx/dx equal to 0 as well?
• Oct 15th 2010, 05:24 PM
Gusbob
Quote:

Originally Posted by lilaziz1
:O! Yes it does!

This seems a bit recursive though. We're basically saying:

y = -3x - z, where z = -(3x+y) and by direct substitution, we get y = -3x + (3x+y) = y

The partial derivative of y = y with respect to x is 0.

In that case, wouldn't dx/dx equal to 0 as well?

It is a bit circular, but you would expect it to be.
After all, F(x,y,f(x,y)) = 3x + y + f(x,y) = 0
describes the exact same thing as f(x,y) = -3x - y
Interchanging the two would seem to be recursive.

As for dx/dx, dx/dx always equals one.
Lets try this again on your example

3x = -y -z

$3\frac{d}{dx} x = 0 - \frac{d}{dx}z = -(-3)$

$\frac{dx}{dx} = 1$

To see what I mean, I'll give a worded illustration. It is easy to get caught up with just the maths and forget why we study it in the first place.

Say we have a container filled with two miscible liquids (two liquids which mix evenly), Suppose z is the amount of liquid flowing out of a container, x be the amount of liquid A flowing out, y be the amount of liquid B flowing out.

Now dy/dz means for every amount dz of total liquid flowing out, a dy amount of liquid B flows out. Similarly, dx/dz is the proportion of liquid A compared to the combined liquid

What does it mean to have dz/dz, dx,dx or dy/dy? For every dz amount of total liquid flowing out, you'll have a dz amount of liquid flowing out. The proportion between total liquid out/ total liquid out is 1.