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Partial Derivatives?

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data

    Okay, so at my school, you can get into Diff EQs without taking Calculus 3. So, I get most of the basis of it, but some things I am missing the boat on.

    How does one go from

    [itex]d(\frac{1}{3}x^3y^3)=x^2y^3dx+x^3y^2dy[/itex] ?

    How do you carry out that operator? It says to take the derivative of that 2 variable function, but does not say with respect to what?

    I am assuming it is some sort of partial derivative...but I am not sure?

  2. jcsd
  3. Mar 3, 2008 #2


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    By definition,


    and df(x,y) is calles the total differential of f.
  4. Mar 3, 2008 #3
    What does [itex]f_x(x,y)dx[/itex] mean? What is f_x?
  5. Mar 3, 2008 #4


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    I've just begun to accept this formula myself.

    The d(stuff) on the left side of the equation contains two variables, x and y. so you take the partial derivative with respect to each of them and add them together.

    f_x simply means the partial function's partial with respect to x. Test it out on the function you've given us.
  6. Mar 3, 2008 #5

    So I take the derivative of [itex]\frac{1}{3}x^3y^3[/itex] first wrt x and then wrt y holding y then x constant, respectively, and then sum the results?
  7. Mar 3, 2008 #6
    So I get [tex]y^3x^2\frac{df}{dy}+x^3y^2\frac{df}{dx}[/tex] then how do I cope with

    the notations? How do I make it look like [itex]x^2y^3dx+x^3y^2dy[/itex]?
  8. Mar 3, 2008 #7


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    You seem to have misunderstood slightly. I re-write what I wrote in post #2:

    By definition,

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


  9. Mar 3, 2008 #8
    I understand how to take the partial of [itex]\frac{1}{3}x^3y^3[/itex] wrt y and x, but I do not understand the NOTATION at all.

    df(x,y) means to take the derivative of f wrt x (hold y as constant) and take the derivative of f wrt y (hold x constant), but what are the resulting notations?
  10. Mar 3, 2008 #9


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    it's like when you do a substitution in Calculus II.

    you have some integral, like

    [tex]\int{\frac{x }{1+x^2}dx}[/tex]

    so you use the substitution [tex]u = x^2[/tex]

    but you have to get the differential for u too, so you take the derivative of both sides of the above equation:

    [tex]du = 2x dx[/tex] (here we have a function u = u(x), where our function above has two variables, so we require the partial derivative formula that quasar provided)

    you could also see it as

    [tex]\frac{du}{dx} = 2x[/tex]

    where you have multiplied by the differential dx, afterwards, treating it like an algebraic quantity (which we can do with practically all differentials in physics... might want to be more careful for a math teacher... maybe justify it)
    Last edited: Mar 3, 2008
  11. Mar 3, 2008 #10
    think about it like in one variable... it is the increment of the function f due to the little dx and the "slope" is f'....right... now u have to coordinate axis x and y...

    more formally the df is a matrix when the function is from R^n----->R^m

  12. Mar 3, 2008 #11
    Okay. Let's keep it simple here guys.
  13. Mar 3, 2008 #12


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    You say you're confused with the notation.

    But you are probably already familiar with the notion of a differential for a function of one variable? Given f differentiable, we define its differential as the function of two variables

    df(x,dx) = f '(x)dx.

    It says "given a value of x and an increment dx, the approximate resulting increment of f is df(x,dx)."

    This is the same thing for the function of two variables. Given a point (x,y) and increments dx and dy in the x and y directions respectively, df(x,y,dx,dy) keeps tract of the approximate resulting increment of f based on the rates of change of f at (x,y) (the partial derivatives)
  14. Mar 3, 2008 #13


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    Try it this way. Suppose x and y were both functions of t. x(t), y(t). Then d/dt(x^3*y^3/3)=x^2*y^3*dx/dt+x^3*y^2*dy/dt. Now suppose they are functions of q, x(q) and y(q). Same thing but with q substituted for t. Big deal. So let's just get it over with and say d(x^3*y^3/3)=x^2*y^3*dx+x^3*y^2*dy. We can divide by d(whatever) when we figure out what the 'whatever' is. It's just a notational thing.
  15. Mar 4, 2008 #14
    :rofl: let's just get it over with... Wonderful. This makes sense. Thanks Dick.
  16. Mar 4, 2008 #15


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    I think you should go to the department chair and tell him/her how foolish it is to let people take partial differential equations without having taken multi-variable calculus first!
  17. Mar 4, 2008 #16
    I never said that I was taking Partial Differential Equations. I am taking Differential Equations. And while defining exactness, some partial derivatives come up.

    I just want to understand their proof fully.
  18. Mar 4, 2008 #17


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    I would consider both Calculus III (Calculus of several varibles) and Linear Algebra necessary prerequisites for "Ordinary Differential Equations".
  19. Mar 4, 2008 #18
    I am no expert, but that hardly seems necessary. But like I said, I am no expert. I would say that it would be nice to have those, but they do not seem requisite of the course content.

    But I may be kicking myself later for saying that! We'll see!
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