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Partial Derivatives

  1. Nov 28, 2014 #1

    Little confused by something:

    if we have u=x+y and v=xy what is the partial derivative w.r.t. u of


    I am told it is 2y (dy/du) = u (dy/du) + y

    And I can see where these terms come from. What I don't understand is why there is no (dv/du) term, as v and u aren't independent.

    Any ideas?
  2. jcsd
  3. Nov 28, 2014 #2


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    The variables x and y have been exchanged with u and v:They are the independent variables, and y is a function of them, y(u,v).
    Any function f(u,v) have partial derivatives with respect to u and with respect to v, but the derivatives of u or v with respect to each other is zero.
  4. Nov 28, 2014 #3
    Definition: A function f is said to be a function of the independent variables (a, b, c) if specifying a, b, c fully determines f (but not overdetermine: i.e. specification of all the three a, b, c are necessary for the determination of f; non-specification of any one of the a, b, c will leave f undetermined). This is our so called 'CONSTRUCTION'. Also, a, b, c are said to be independent variables.

    Now, take y^2 be z. Then z=z(u, v, y) - as per the definition above, and all three, i.e. u, v, y are independent (as per above). Don't go by 'u=x+y and v=xy' to conclude that u and v aren't independent.

    Also, I don't think it's correct to write y=y(u,v), since specification of u, v doesn't fully determine y (it's a quadratic equation-hence two roots)
    But, y^2=y^2(u, v, y) is correct, since y^2 is fully determined by u, v and y.
  5. Nov 29, 2014 #4


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    is it not overdetermined????
  6. Nov 29, 2014 #5
    I see. True.
    By the way, can we even legitimately talk of partial derivative of y^2=uy-v wrt u? Before doing that, we I think we should decide our independent variables on which y^2 depends.
    Also, in
    'dy/du' indicates a total derivative, so y should be fully determined by u. But we see, it's not.
  7. Nov 29, 2014 #6

    Stephen Tashi

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    We could equally well ask: What is the partial derivative of y^2 with repsect to v ?

    This general type of confusion often occurs in physics problems. Authors establish a complicated relation between several variables and then write a total differential that seems to be missing some partial derivatives. I conjecture the problem lies in ambiguous notation. The notation y^2 = uy - v does not specify the function y^2 as being a function of particular variables. A correct mathematical definition of a function states a particular domain for a function. A relation like y^2 = uy - v does not.

    The "given" information to determine a value of a mathematical function is specified by a vector of values for specific variables. The value of the left hand side of an equation might be determined from a variety of information about the right hand side.
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