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Partial Derivatives

  1. Apr 21, 2008 #1
    http://img132.imageshack.us/img132/5736/wathh1.jpg

    I think to find dz/dx (d = delta) first of all is by

    dz/dx = (dz/du)(du/dx) + (dz/dv)(dv/dx)

    But how do I find dz/du and dz/dv for this?

    I only have 1 example that resembles this and it had z defined as a function as z = u^v, but here I dont have that, so how can I find it?

    Thanks
     
  2. jcsd
  3. Apr 21, 2008 #2
    find dz/du * du/dx - take the derivative of u with respects to x
     
  4. Apr 21, 2008 #3
    Sorry I don't understand, I'm trying to find dz/du first then dz/dx.

    If I take the derivative of u with respect to x and hold y constant, then i get the partial derivative of du/dx. Don't I?
     
  5. Apr 21, 2008 #4
    I don't think you mean delta. I think that you're writing the chain rule and meant to write partial derivatives and the d's are [tex]\partial[/tex] right?

    I'll give you an example:

    [tex]
    z = uv, u=\cos (x+y), v=\sin (x+y)
    [/tex]
    then clearly [tex]z=\frac{1}{2}\sin2(x+y) \Rightarrow \frac{\partial z}{\partial x} = \cos2(x+y)[/tex]

    But now let's use your formula and see if we get the same thing:
    [tex]\frac{\partial z}{\partial u}=v=\sin(x+y),\frac{\partial u}{\partial x}=-\sin(x+y)[/tex]
    [tex]\frac{\partial z}{\partial v}=u=\cos(x+y),\frac{\partial v}{\partial x}=\cos(x+y)[/tex]

    So that
    [tex]\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y} \Rightarrow[/tex]
    [tex]\frac{\partial z}{\partial x}=-\sin(x+y)\sin(x+y)+\cos(x+y)\cos(x+y)[/tex]
    [tex]\frac{\partial z}{\partial x}=\cos2(x+y)[/tex]

    And hey it's the same thing! So in that example all the derivatives are partial derivatives, and you keep the other variables constant when you evaluate them.
     
  6. Apr 21, 2008 #5
    Ah sorry ye I thought the curly d was a delta :P
     
  7. Apr 21, 2008 #6
    Thanks for your reply, so am I right in assuming dz/du = v and dz/dv = u?

    I didn't know that, thanks.
     
  8. Apr 21, 2008 #7
    Not in general, just in that example. For instance if you have [tex]z=u^2v^3[/tex] then [tex]\frac{\partial z}{\partial u}=2uv^3[/tex] and [tex]\frac{\partial z}{\partial v}=3uv^2[/tex].
     
  9. Apr 21, 2008 #8
    ok well in my question

    http://img132.imageshack.us/img132/5736/wathh1.jpg

    I plan on finding dz/du and dz/dv, but how do I go about this?

    I don't have a z = .. like you are using in your examples unless I am misunderstanding the notation in the question.

    I then plan on spending a while figuring out where to go from there :)

    Edit: It says z = f(x,y) but I'm not sure what this means exactly.
     
    Last edited: Apr 21, 2008
  10. Apr 21, 2008 #9
    :/ I just found more examples in a text book, they all have f(x,y) equal to something, here I do not, is it a mistake with the question?

    Its only worth 4 marks and it's taking me forever to do.
     
  11. Apr 21, 2008 #10

    Hootenanny

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    I'm assuming that your function f is complex valued and therefore may be written,

    [tex]z = f(x,y) = u + i v = x+ay + i\left(x-ay\right)[/tex]

    Therefore,

    [tex]\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + i\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}[/tex]

    [tex] \Rightarrow z_x = \frac{\partial}{\partial x}\left(x+ay\right) + i\frac{\partial}{\partial x}\left(x-ay\right)[/tex]

    An similarly for partial derivatives with respect to y. Does that help?
     
    Last edited: Apr 21, 2008
  12. Apr 21, 2008 #11
    Yes thankyou, I'll work on from there :)
     
  13. Apr 21, 2008 #12
    I am using Glyn James, and I had like 20 these like questions! :D (almost in every single exercise related to partial ...), and also we always get this type of question on the exam:

    Here's two good questions:
    1)
    u = xy
    v = x-y
    z = f(u,v)

    show :
    x.dz/dx - y.dz/dy = (x-y)* (dz/du)

    2)(little harder)
    u = x+ay
    v = x+by
    z = f(u,v)

    show
    9 (d^2f/dx^2) - 9 (d^2f/dx.dy) + 2 (d^2f/dy^2) = 0
    into
    d^2f / du.dv = 0


    Main strategy to solve these questions: draw a tree with vars!
     
  14. Apr 21, 2008 #13
    Hootenanny, your solution does not any sense to me >>

    Here's how I think it should be done:
    z = f(x,y)
    x = u+v
    y = (u-v)/a

    dz/du = dz/dx . dx/du + dz/dy. dy/du
    -Applying multiplication rule ..
    d^2z / dv.du = d/dv (dz/dx . dx/du ) + ...
    = [d^z/dx^2 . (dx/du)^2 + d^2z /dx.dy * dx/du * dy/du ]+ dz/dx * dx/ du.dv * dv/dx ...

    wow, it so messy but it should go like that if not necessarily exactly like above
    and you final answer may have some deltas in it
     
  15. Apr 21, 2008 #14

    Hootenanny

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    z is usually used to denote complex variables, hence I assumed that f(x,y) was complex valued. If f is indeed complex valued then u and v simply represent the real and imaginary parts of the function and the function can be differentiated as I have shown above.
     
  16. Apr 21, 2008 #15
    ook thnx.
    They never told/made us think that way><
    [But thinking either way shouldn't give different answer, right?]
     
  17. Apr 13, 2009 #16

    Lyn

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    Hi i am struggling too with this same question, http://img132.imageshack.us/img132/5736/wathh1.jpg

    I have tried to follow what you have said and i get my (d's are partial derivatives) dz/dx = 2. I am unsure if this is correct because just like the problem the other person had with this question, i don't have a z=.... (eg z=xy)
     
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