1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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]

    [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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook