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Total derivative: elementary question

  1. Feb 3, 2010 #1
    Given a function [tex]f(x_1,\ldots,x_n)[/tex], the total derivative for [itex]x_1[/itex] is:

    [tex]\frac{df}{dx_1}=\frac{\partial f}{\partial x_1}+\frac{\partial f}{\partial x_2}\frac{dx_2}{dx_1}+\ldots+\frac{\partial f}{\partial x_n}\frac{dx_n}{dx_1}[/tex]

    Now, if the [itex]x_i[/itex] are just variables, can we say that:


    ?...If so, then the total derivative would equal the partial derivative:

    [tex]\frac{df}{dx_1}=\frac{\partial f}{\partial x_1}[/tex]

    is that correct?
  2. jcsd
  3. Feb 3, 2010 #2
    Correct. Total derivatives are only used when the parameters of a function are not independent. This is really a bastardization of the term function, since all formal parameters are independent by definition. But we let the physicists do it anyway, because they're so swell.

    It makes more sense, I think, if you parametrize all the variables over one parameter (say t):

    [tex]f(x_1(t), x_2(t), \dots, x_n(t))[/tex]

    And then take the regular Calc I derivative with respect to t

    [tex]\frac{d}{dt} f(x_1(t), x_2(t), ..., x_n(t)) = \frac{\partial f}{\partial x_1}\frac{dx_1}{dt} + \frac{\partial f}{\partial x_2}\frac{dx_2}{dt} + \dots + \frac{\partial f}{\partial x_n}\frac{dx_n}{dt}[/tex]
  4. Feb 3, 2010 #3
    Thanks a lot!
    Your explanation was very clear.
    I have just got something which is still bugging my mind.

    What about the following example:



    we want to compute the total derivative [tex]\frac{df}{du}[/tex].
    This would yield:

    [tex]\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}\frac{dv}{du}[/tex]

    Now, what about [tex]\frac{dv}{du}[/tex]. Is it supposed to be zero???
  5. Feb 3, 2010 #4
    It occurs to me that I made a slight mistake in my explanation above. You can see on the RHS of my last equation, I'm missing a [tex]\frac{\partial f}{\partial x}[/tex] term, so it's not exactly equal to the total derivative except in the case where x1(t) = t.

    You can see how tricky these problems can get!

    In your second example, I'm not sure the best way to interpret it. Do you realize you're using the variable x twice?

    Anyway, the moral in the story is these problems are often written with a particular physical set up in mind, and the details are only summarized by the equations used. Numbers get mistaken for FUNCTIONS of real numbers, and it is impossible to attack the problem in an algebraic way. You have to reason about the problem and use techniques like the total derivative which are specially designed to minimize the negative effects of the notation used in calculus.
  6. Feb 3, 2010 #5
    I can't see any mistake in your explanation!
    Why is that?
    Also wikipedia http://en.wikipedia.org/wiki/Total_derivative reports the same example you mentioned: it is at the and of the section "Differentiation with indirect dependencies"

    I agree with you that these things are tricky!

    I suspect the tricky example I made arises when you deal with curvilinear coordinates.
    Some time ago I had a problem which I thought I understood, but I am afraid I still don't.
    The problem is explained in this thread: https://www.physicsforums.com/showthread.php?t=365940

    I hope you don't mind if I ask you to take a look at it. But in case you have no time, thanks anyway!
  7. Feb 3, 2010 #6

    Ben Niehoff

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    This is correct. The assumption here is that the x_i might have hidden dependencies between them. This sort of situation occurs in, e.g. thermodynamics.

    This is not true at all. If you use partial derivatives,

    [tex]\frac{\partial x_i}{\partial x_j} = \delta_{ij}[/tex]

    then it is true. But if the x_i have hidden dependencies between them, then the total derivative will not be trivial.

    Not at all, that makes no sense. I hope the above explanation helps you see why.
  8. Feb 3, 2010 #7
    I think there is a fundamental gap in my knowledge.
    Perhaps I don't fully understand the meaning of [itex]dx_2/dx_1[/itex] when they are two variables.

    Let's make a very simple example:


    the total derivative is:

    [tex] \frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx} = 1 + \frac{dy}{dx}[/tex]

    Now, what do we replace [tex]\frac{dy}{dx}[/tex] with???
    This notation assumes that y is a function of x, which is not true!
    Last edited: Feb 3, 2010
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