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    The question is in the attached file. Thanks for any help offered.
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  1. File Type: doc If z.doc (40.0 KB, 93 views)
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    Are you sure it's not 3x(dz/dx)= 2y(dz/dy)?
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    (Original post by EierVonSatan)
    Are you sure it's not 3x(dz/dx)= 2y(dz/dy)?
    Err..no, it is partial differential as stated. The f does simply means that z is a function of x^2y^3 and therefore z is a function of x and y right?
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    (Original post by shengoc)
    Err..no, it is partial differential as stated. The f does simply means that z is a function of x^2y^3 and therefore z is a function of x and y right?
    yes, I know its partial (I'm being lazy) my question was to the order of the derivatives...
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    (Original post by EierVonSatan)
    yes, I know its partial (I'm being lazy) my question was to the order of the derivatives...
    meaning...
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    (Original post by shengoc)
    meaning...
    is it 2y\frac{\partial z}{\partial x} = 3x\frac{\partial z}{\partial y} or 3x\frac{\partial z}{\partial x} = 2y\frac{\partial z}{\partial y}?

    As I can't get z = x2y3 to fit the former.
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    (Original post by EierVonSatan)
    Are you sure it's not 3x(dz/dx)= 2y(dz/dy)?
    I think it should be this - thiss works.

    When it is 3x(pz/py) = / = 2y(pz/px)

    However,

    3x(pz/px) = 2y(pz/py)

    The question itself is wrong.
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    (Original post by DeanK2)
    I think it should be this - thiss works.

    When it is 3x(pz/py) = / = 2y(pz/px)

    However,

    3x(pz/px) = 2y(pz/py)

    The question itself is wrong.
    what is that p you used in your equation anyway?
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    (Original post by shengoc)
    what is that p you used in your equation anyway?
    p = partial symbol ...I would imagine
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    (Original post by EierVonSatan)
    is it 2y\frac{\partial z}{\partial x} = 3x\frac{\partial z}{\partial y} or 3x\frac{\partial z}{\partial x} = 2y\frac{\partial z}{\partial y}?

    As I can't get z = x2y3 to fit the former.
    No, EVS, the question was stated that way, and it is 3x(partial of z wrt y at constant x) = 2y(partial of z wrt x at constant y), just to make sure i didn't type anything wrong in the file i attached.

    But yeah, am I allowed to imply that z = x^2 x y^3 ?
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    It seems from the document that all you have been asked to show is the scanned equation holds true for a function z you are free to choose.

    It looks like the statemnt if z = (x^2)(y^3) is written by you.

    Why don't you try a different function for z?
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    Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u2 - 1 and let u = x^2y^3

    knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

    Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

    so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

    but 3x(4x^3y^6)= 2y(6x^4y^5) ?
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    (Original post by EierVonSatan)
    Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u2 - 1 and let u = x^2y^3

    knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

    Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

    so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

    but 3x(4x^3y^6)= 2y(6x^4y^5) ?
    I see your point, but i was told that what we are asked to show in this question can be done. Hmm, thanks anyway, for all your help.
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    (Original post by EierVonSatan)
    Taking a random example (as the relation should hold for all 'nice' functions) f(u) = u2 - 1 and let u = x^2y^3

    knowing that x and y are independant (i.e. z = f(x, y) =/= 0)

    Partial (dz/dx) = 4x^3y^6 and partial (dz/dy) = 6x^4y^5

    so 2y(4x^3y^6) =/= 3x(6x^4y^5) according to the relation

    but 3x(4x^3y^6)= 2y(6x^4y^5) ?
    Yes, i managed to show what the question ask to show. Apparently, a chain rule works like you mentioned above. If I let z=f(u), u = x^2 y^3

    therefore dz/dx = dz/du x du/dx

    i did the same for dz/dy,

    then make dz/du as subject of formula and equate the two, then i managed to show it. Does this make sense to you?

    p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS
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    (Original post by shengoc)
    p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS
    You do realise that this is what I was asking you from the beginning? :rolleyes:
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    (Original post by shengoc)
    Yes, i managed to show what the question ask to show. Apparently, a chain rule works like you mentioned above. If I let z=f(u), u = x^2 y^3

    therefore dz/dx = dz/du x du/dx

    i did the same for dz/dy,

    then make dz/du as subject of formula and equate the two, then i managed to show it. Does this make sense to you?

    p/s - I just realised there may be errors in the question, i got 3x(partial z wrt x) in the end instead of what the question stated. hmm, i will leave it that way then, thanks a lot, EVS
    There are bound to be errors in the question - it's Grout.
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    (Original post by EierVonSatan)
    You do realise that this is what I was asking you from the beginning? :rolleyes:
    no, actually i don't; i didn't have a clue how to do that until the chain rule comes up and only then, i realise that, my apologies for that, EVS.
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    (Original post by cpchem)
    There are bound to be errors in the question - it's Grout.
    Thanks, yeah, i checked all my workings again, and it seemed to work out logically, if the question was set otherwise. Anyway, thanks for letting me know.
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    (Original post by shengoc)
    no, actually i don't; i didn't have a clue how to do that until the chain rule comes up and only then, i realise that, my apologies for that, EVS.
    Not your fault really :p:
 
 
 
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