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Homework Help: Partial derivative

  1. Nov 2, 2014 #1
    1. The problem statement, all variables and given/known data
    If w = w(x, y, z) is given implicitly by F(x, y, z, w) = 0, find a formula
    for both ∂w/∂z and ∂^2w/∂y∂z . You may assume that each function is sufficiently
    differentiable and anything you divide by during the process of your
    solution is non-zero.

    3. The attempt at a solution
    I know in class, my professor showed us how to implicitly differentiate using this method where if y = y(x) and F(x,y) = 0, you can do dF/dx = ∂F/∂x + ∂F/∂y*dy/dx = 0 and dy/dx = -(∂F/∂x)/(∂F/∂y),
    Is this the kind of formula my professor wants us to find?
  2. jcsd
  3. Nov 2, 2014 #2
    You have to use the chain rule..
  4. Nov 2, 2014 #3
    [itex]0 = \frac{\partial F}{\partial x} \frac{\partial x}{\partial z} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial z} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial z} + \hspace{1 mm}...[/itex]
    Last edited: Nov 2, 2014
  5. Nov 2, 2014 #4
    Yea I got ∂w/∂z, but I have no idea how to get the other one, do I do ∂/∂y(∂w/∂z)?
  6. Nov 2, 2014 #5
  7. Nov 2, 2014 #6
    I am not sure what ∂/∂y(∂w/∂x*∂x/∂z) would be, would it just be, d^2w/∂x∂y * ∂^2x/∂z∂y?
  8. Nov 2, 2014 #7
    Show us what you got for ##\partial w/\partial z##, and then we can discuss what to do next.

  9. Nov 2, 2014 #8
    I dont think so, I believe you'll need to apply the product rule to the expression that you used to get your first answer..
  10. Nov 2, 2014 #9
    0 = ∂F/∂x + ∂F/∂y + ∂F/∂z + ∂F/∂w*∂w/∂z

    ∂w/∂z = -(∂F/∂x + ∂F/∂y + ∂F/∂z)/∂F/∂w

    Looks like I might needs quotient rule.
  11. Nov 3, 2014 #10
    It looks like it needs more than just that. This result is incorrect.

  12. Nov 3, 2014 #11


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    Look at post number 3.
    Usually it is safe to assume that ##\partial x / \partial z =0##.
  13. Nov 3, 2014 #12
    Especially if x is one of the parameters that is being held constant.
  14. Nov 3, 2014 #13
    I don't know what to do at all, I am doing exactly what my prof said to do, make a tree, with F at the top then branch out the different variables then branch out w since it is a function of x,y,z. What exactly did I do wrong in my answer?
  15. Nov 3, 2014 #14
    I think that you just failed to realize that:
    [itex] \frac{\partial F}{\partial x} * 0 = 0 [/itex]...there are some terms in your answer that should not be there...
  16. Nov 3, 2014 #15


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    The first three terms of the original differentiation are missing factors. Presumably what you want is
    0=\frac{\partial F}{\partial z}=
    \frac{\partial F}{\partial x}\frac{\partial x}{\partial z}+
    \frac{\partial F}{\partial y}\frac{\partial y}{\partial z}+
    \frac{\partial F}{\partial z}\frac{\partial z}{\partial z}
    \frac{\partial F}{\partial w}\frac{\partial w}{\partial z}$$
    Then use the fact that
    $$\frac{\partial x}{\partial z}=
    \frac{\partial y}{\partial z}=0$$
    $$\frac{\partial z}{\partial z}=1$$
  17. Nov 3, 2014 #16
    Why are you multiplying them by ∂x/∂z and ∂y/∂z? If you don't mind me asking, I was never taught you had to do that, well maybe I was but I don't remember, I am looking through my notes and it wasn't mentioned. Like I understand with ∂F/∂w, but not the others.
  18. Nov 3, 2014 #17
    It is because you are required to use the chain rule, you are differentiating a composite function of more than one variable..
  19. Nov 3, 2014 #18
    Maybe a (somewhat) unrelated example may clear things up:
    Suppose you have a function [itex] z = f(x,y) [/itex] where [itex] x = g(s,t) [/itex] and [itex] y=h(s,t) [/itex]. If you want to find the derivative of z with respect to s (a similar situation to the problem you are trying to solve) then you would use the chain rule as follows:
    [itex] \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} [/itex]
  20. Nov 3, 2014 #19
    Oh alright I think I understand looking at the definition, but in class all the examples were like where x,y,z where all functions of t, or s and t, and I never realized that if they aren't a function of anything you still have to use chain rule on it.
  21. Nov 3, 2014 #20
    You have to use the chain rule because you have:
    [itex] w = f(x,y,z) [/itex] given implicitly by [itex] F(x,y,z,w) = 0 [/itex]. So you essentially have [itex] F(x,y,z, f(x,y,z)) = 0 [/itex]. Which is a composite function.
  22. Nov 3, 2014 #21


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    The difference between curly and straight 'd's is a bit tricky in that one. What I wrote above doesn't get them correct because ##\frac{\partial F}{\partial z}## implies that ##w##, as well as ##x,y##, is kept constant, which prevents us from using the fourth term in the way we want. So I think we need to write it as a total derivative:

    \frac{\partial F}{\partial x}\frac{dx}{dz}+
    \frac{\partial F}{\partial y}\frac{dy}{dz}+
    \frac{\partial F}{\partial z}\frac{dz}{dz}
    \frac{\partial F}{\partial w}\frac{dw}{dz}$$

    This then gives us an answer for ##\frac{dw}{dz}## rather than ##\frac{\partial w}{\partial z}##.

    But that's OK because

    \frac{\partial w}{\partial x}\frac{dx}{dz}+
    \frac{\partial w}{\partial y}\frac{dy}{dz}+
    \frac{\partial w}{\partial z}\frac{dz}{dz}=0+0+1\times\frac{\partial w}{\partial z}$$
  23. Nov 3, 2014 #22
    Let's take a step back. Are you familiar with the following?
    [tex]dF=\left(\frac{\partial F}{\partial x}\right)_{y,z,w}dx+\left(\frac{\partial F}{\partial y}\right)_{x,z,w}dy+\left(\frac{\partial F}{\partial z}\right)_{x,y,w}dz+\left(\frac{\partial F}{\partial w}\right)_{x,y,z}dw[/tex]
    But, since F is zero, dF = 0, so
    [tex]\left(\frac{\partial F}{\partial x}\right)dx+\left(\frac{\partial F}{\partial y}\right)dy+\left(\frac{\partial F}{\partial z}\right)dz+\left(\frac{\partial F}{\partial w}\right)dw=0[/tex]
    [tex]dw=-\frac{\partial F/\partial x}{\partial F/\partial w}dx-\frac{\partial F/\partial y}{\partial F/\partial w}dy-\frac{\partial F/\partial z}{\partial F/\partial w}dz[/tex]

    Compare this term for term with
    [tex]dw=\left(\frac{\partial w}{\partial x}\right)dx+\left(\frac{\partial w}{\partial y}\right)dy+\left(\frac{\partial w}{\partial z}\right)dz[/tex]

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