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Total derivative?

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Is the following equation a total derivative?

    [tex] dz = 2ln(y)dx+{\frac x y}dy[/tex]

    2. Relevant equations
    -


    3. The attempt at a solution

    I would say no. I tried it with the symmetry of the second derivatives.

    [tex] 2ln(y) [/tex] is [tex] {\frac {\partial z} {\partial x}}[/tex]

    when I derivate it now and say y is my variable I get:

    [tex] {\frac {\partial z} {\partial y}}=\frac 2 y[/tex]

    same for the other expression

    [tex] {\frac {\partial z} {\partial y}}=\frac x y[/tex]

    [tex] {\frac {\partial z} {\partial x}}=\frac 1 y[/tex]


    so they are different => no totale derivative. Am I right or where did I do mistakes?

    Thanks for the help
     
  2. jcsd
  3. Jun 28, 2011 #2

    tiny-tim

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    Hi Lindsayyyy! :wink:

    That's completelyyyy right. :smile:

    (just to check, I like to differentiate the obvious possibility, xlny, and that gives lnydx + xdy/y, which doesn't match)
     
  4. Jun 28, 2011 #3
    Thank you very much:smile:
     
  5. Jun 28, 2011 #4

    HallsofIvy

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    Just to put my oar in, if that were a total derivative, there would exist a function, F(x,y) such that
    [tex]\frac{dF}= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy[/tex]
    [tex]\frac{dF}= 2 ln(y)dx+ \frac{x}{y}dy[/tex]
    ..
    So we must have [itex]\partial F/\partial x= 2 ln(y)[/itex] so [itex]F= 2x ln(y)+ \phi(y)[/itex] where [itex]\phi(y)[/itex] can be any differentiable function of y only.

    From that [itex]\partial F/\partial y= 2x/y+ \phi'(y)[/itex] and that must be equal to [itex]x/y[/itex]: [itex]2x/y+ \phi'(y)= x/y[/itex] so [itex]\phi'(y)= -x/y[/itex] which is impossible since [itex]\phi'(y)[/itex] is not a function of y.

    Note the "mixed second derivatives". If [itex]\partial F/\partial y= x/y+ \phi'(y)[/itex]
    then [itex]\partial^2 F/\partial x\partial y= 1/y[/itex] and if [itex]\partial F/\partial x= 2ln(y)[/itex] then [itex]\partial^2 F/\partial y\partial x= 2/y[/itex]. Those mixed second derivatives should be the same and that was what you were checking.
     
  6. Jun 28, 2011 #5
    your solution is way more elegant than mine :smile:

    I have another problem:

    [tex]x^{\alpha}y^{\beta}(2ln(y)dx+{\frac x y}dy)[/tex] is given. I shall find alpha and beta that the given expression is a total derivative.

    My approach:

    I found [tex]{ \frac {\partial^{2}z} {\partial x \partial y}}=2\beta x^{\alpha}y^{\beta -1}ln(y)+{\frac 1 y}2x^{\alpha}y^{\beta}[/tex]

    and
    [tex]{ \frac {\partial^{2}z} {\partial y \partial x}}=\alpha x^{\alpha -1}y^{\beta}{\frac x y}+{\frac {x^{\alpha}y^{\beta}} y} [/tex]

    I though I could equate them and solve the equation to get an expression for alpha and beta. But my problem is I can easily cancel some terms with potential rules like y^a-1=y^a/y etc.

    And I get to the euqation:

    [tex] 2\beta lny+2 = \alpha+1[/tex]

    is my approach wrong, where are my mistakes? If I was right thus far, how should I go on?

    Thanks for your help
     
  7. Jun 28, 2011 #6

    SammyS

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    There is no y on the right hand side, so what must β be ?
     
  8. Jun 28, 2011 #7
    fail :yuck:

    [tex] \beta =0 \alpha =1 [/tex]

    thanks, I'm blind, I know it's a bit too much and insolent to ask but you haven't calculated it by yourself in order to test if I was right until the last step, have you?

    Thanks for your help

    edit:

    nevermind, I'm tired^^
    I haven't thought about just to put it in and derive it haha. It works. thanks
     
    Last edited: Jun 28, 2011
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