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Math Help - Implicit Function Theorem

  1. #1
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    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.

    Thanks in advance!
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    Quote Originally Posted by lilaziz1 View Post
    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.

    Thanks in advance!
    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))
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  3. #3
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    Quote Originally Posted by lilaziz1 View Post
    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.

    Thanks in advance!

    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
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    Quote Originally Posted by Gusbob View Post
    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?
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    Quote Originally Posted by lilaziz1 View Post
    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?
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  6. #6
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    :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?
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  7. #7
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    Quote Originally Posted by lilaziz1 View Post
    :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.
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