Exact differential

  1. how would i show that this is the exact differential

    z=xy-y+lnx+2
    dz= (y+ 0)dx + (x-1)dy (i hope the total differential is right)

    so how would i show that dz is the exact differential ???
     
  2. jcsd
  3. quasar987

    quasar987 4,774
    Science Advisor
    Homework Helper
    Gold Member

    The definition of the total differential of a function f(x,y) is the expression

    [tex]df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy[/tex]

    Now with the z you're given, is the expression for dz correct? That is to say, is

    [tex]\frac{\partial z}{\partial x}=(y+0)[/tex]

    and

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

    ??
     
  4. i believe dz is correct because i used partial differentials to solve it, and i hope its correct
     
  5. quasar987

    quasar987 4,774
    Science Advisor
    Homework Helper
    Gold Member

    what is the derivative of ln(x)?
     
  6. HallsofIvy

    HallsofIvy 40,785
    Staff Emeritus
    Science Advisor

    Once you fixed the derivative of ln x thing, the fact that dz is the total derivative of z pretty much means it is an "exact" differential. That's the definition of "exact" differential!
     
  7. thx and the derivative of lnx is just 1/x
     
  8. HallsofIvy

    HallsofIvy 40,785
    Staff Emeritus
    Science Advisor

    Yes.
    z=xy-y+lnx+2 so
    dz= (y+ 1/x)dx + (x-1)dy

    And that is an "exact" differential precisely because it is the differential of z.

    Now suppose you were given the differential form (y+ 1/x)dx+ (x- 1)dy without having been given z. How would you determine whether it was an exact differential?
    If
    [tex](y+ 1/x)= \frac{\partial z}{\partial x}[/tex]
    for some function z and
    [tex]x-1= \frac{\partial z}{\partial y}[/tex]
    then
    [tex]\frac{\partial(y+ 1/x)}{\partial y}= \frac{\partial^2 z}{\partial x\partial y}[/tex]
    and
    [tex]\frac{\partial(x-1)}{\partial x}= \frac{\partial^2 z}{\partial y\partial x}[/tex]
    and so must be equal. Fortunately, it is easy to see that each second derivative is 1 and so they are in fact equal.
    Okay, now, how would we find z? Knowing that
    [tex]x-1= \frac{\partial z}{\partial y}[/tex]
    integrating with respect to y (treating x as a constant) we get z= xy- y+ C, except that, since we are treating x as a constant, that "constant of integration", C, may depend on x: z= xy- y+ C(x). Differentiating that with respect to x,
    [tex]\frac{\partial z}{\partial x}= y+ C'(x)= y+ 1/x[/tex]
    Notice that the "y" terms cancel (it was the "cross condition" above that guarenteed that) and so we have C'(x)= 1/x. Then C(x)= ln(x)+ C where "C" now really is a constant.
    Any z(x,y)= xy- y+ ln(x)+ C satisfies dz= (y+ 1/x)dx+ (x-1)dy.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?