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Partial derivative

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data
    what s the wrong with the following arguments
    suppose that w=f(x,y)and y=x^2
    by the chain rule
    (for partial derivative )
    Dw/Dx=(Dw/Dx)( Dx/Dx)+(Dw/Dy)(Dy/Dx)=Dw/Dx+2x( Dw/Dy)
    hence 2x( Dw/Dy)=0 ,and so Dw/Dy=0

    2. Relevant equations

    3. The attempt at a solution
    i think the argument is not right because we can not write Dx/Dx and w has the relationship with x and y ,so we don't need to use Dy/Dx Dx/Dx to get x ,we can do it directly
    is there any counterexample ?
    Last edited: Mar 19, 2009
  2. jcsd
  3. Mar 19, 2009 #2


    Staff: Mentor

    You can't write dw/dx because f is a function of two variables. The derivatives you have to work with are
    [tex]\frac{\partial w}{\partial x}[/tex]
    [tex]\frac{\partial w}{\partial y}[/tex]

    I don't see that it makes any difference that y happens to be equal to x^2. If both x and y were functions of a third variable, say t, then you could talk about dw/dt, but to get it you would still need both partial derivatives and would need to use the chain rule.
  4. Mar 19, 2009 #3
    all of them are partial derivative ,because i don't know how to type the symbol for partial derivative
  5. Mar 19, 2009 #4


    Staff: Mentor

    I can't see any justification for expanding
    [tex]\frac{\partial w}{\partial x}[/tex]
    the way you did. Certainly
    [tex]\frac{\partial w}{\partial x} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial x}[/tex]
    since the latter partial derivative is 1, but I don't see any way that you can add all the other stuff (i.e., the other partials).

  6. Mar 19, 2009 #5
    the problem is that i have to prove the argument is false
  7. Mar 19, 2009 #6


    Staff: Mentor

    Come up with a counterexample for which the statement isn't true. Pick a function of two variables f(x, y), such as f(x, y) = 3x + 2y, where y = x^2.
  8. Mar 19, 2009 #7
    thx for your help
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