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Is there such a thing as a total partial derivative?

  1. Jun 25, 2013 #1
    Is there such a thing as a total "partial" derivative?

    Total Derivative as I've Been Taught
    From my understanding, if we have a function s = f(x, y) where the two arguments x and y are related by another function y = g(x), then there is a great deal of difference between ds/dx and ∂s/∂x.

    ∂s/∂x is simply a partial derivative and can be calculated by treating y as a constant and differentiating f(x, y) with respect to x.

    On the other hand, the "total derivative" ds/dx takes the y = g(x) relationship into account and, by the Chain Rule, gives:
    [itex]\frac{ds}{dx}[/itex] = [itex]\frac{∂s}{∂x}[/itex][itex]\frac{dx}{dx}[/itex] + [itex]\frac{∂s}{∂y}[/itex][itex]\frac{dy}{dx}[/itex]​

    This approach is very well explained in Wikipedia:
    http://en.wikipedia.org/wiki/Total_derivative#Differentiation_with_indirect_dependencies

    A Different Case
    However, what happens if we have a function s = f(x, y, z) and only two of the arguments are related, as through y = g(x).

    As before, ∂s/∂x can still be calculated by differentiating f(x, y, z) and treating y and z as constants, but what of the total derivative in terms of x?

    Such a total "partial" derivative would take the form:
    (total partial derivative in terms of x) = [itex]\frac{∂s}{∂x}[/itex][itex]\frac{dx}{dx}[/itex] + [itex]\frac{∂s}{∂y}[/itex][itex]\frac{dy}{dx}[/itex]​

    But, clearly, we can't notate this as ds/dx since s is also a function of z. Neither can we call is ∂s/∂x since that notation is reserved for the regular partial derivative.

    So my question: is there such a concept as a total "partial" derivative"? I haven't been able to find any discussion on such a concept and was curious about whether something like this even exists.

    Any replies are appreciated, and thank you in advance!
     
    Last edited: Jun 25, 2013
  2. jcsd
  3. Jun 25, 2013 #2
    [itex]\frac{ds}{dx}[/itex] = [itex]\frac{∂s}{∂x}[/itex] + [itex]\frac{∂s}{∂y}[/itex][itex]\frac{dy}{dx}[/itex]+ [itex]\frac{∂s}{∂z}[/itex][itex]\frac{dz}{dx}[/itex]
    If [itex]\frac{dz}{dx}[/itex]=0 then [itex]\frac{ds}{dx}[/itex] = [itex]\frac{∂s}{∂x}[/itex] + [itex]\frac{∂s}{∂y}[/itex][itex]\frac{dy}{dx}[/itex]
     
  4. Jun 25, 2013 #3
    If [itex]\frac{dz}{dx}[/itex]=0, why would we refer to the quantity on the right as [itex]\frac{ds}{dx}[/itex]?

    This is STILL only a partial derivative of s since s depends on z as well.
     
  5. Jun 25, 2013 #4
    I think you are kidding !
     
  6. Jun 25, 2013 #5
    Pardon? Forgive me if I'm missing something.

    s depends on x, y, and z.
    x and y are related; z is independent of the two.

    Thus, considering the relation between x and y, we can think of s as a function of x and z only.
    s then has two PARTIAL derivatives ∂s/∂x and ∂s/∂z.
     
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