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Partial differentiation

  1. Feb 25, 2006 #1
    Say, E is dependent to x,y,z. I'm expecting it's derivative at [tex]x_0,y_0,z_0[/tex] to be

    [tex]dE = \lim_{\substack{\Delta x\rightarrow 0\\\Delta y\rightarrow 0\\\Delta z\rightarrow 0}} E(x_0+\Delta x, y_0+\Delta y,z_0+\Delta z) - E(x_0,y_0,z_0)[/tex]

    But with following definition, it's not the thing above:

    [tex]dE = \lim_{\substack{\Delta x\rightarrow 0\\\Delta y\rightarrow 0\\\Delta z\rightarrow 0}} \frac{E(x_0+\Delta x, y_0,z_0) - E(x_0,y_0,z_0)}{\Delta x} \Delta x + \frac{E(x_0, y_0+\Delta y,z_0) - E(x_0,y_0,z_0)}{\Delta y} \Delta y + \frac{E(x_0, y_0,z_0+\Delta z) - E(x_0,y_0,z_0)}{\Delta z} \Delta z = \frac{\partial E}{\partial x}dx + \frac{\partial E}{\partial y}dy + \frac{\partial E}{\partial z}dz[/tex].

    Now, which is correct? (and why?!?)
    Last edited: Feb 25, 2006
  2. jcsd
  3. Feb 25, 2006 #2


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    Neither one. The differential (not "derivative") is
    [tex]dE= \left(\lim_{\Delta x\rightarrow0}\frac{E(x_0+\Delta x,y_0,z_0)-E(x_0,y_0,z_0)}{\Delta x}\right)dx+ \left(\lim_{\Delta y\rightarrow 0}\frac{E(x_0,y_0+\Delta y,z_0)-E(x_0,y_0,z_0)}{\Delta y}\right)dy+ \left(\lim_{\Delta z\rightarrow 0}\frac{E(x_0,y_0,z_0+\Delta z)-E(x_0,y_0,z_0)}{\Delta z}\right)dz[/tex]
    That's the same as your second formula except for the "d" rather than "[itex]\Delta[/itex]" in the numerator.

    Why would you "expect" the first? In one variable, would you "expect" the differential of f(x) to be
    [tex]df= lim_{\substack{\Delta x\rightarrow 0}}f(x_0+\Delta x)- f(x_0)[/tex]
    If f is any continuous function, that's 0!
    Last edited: Feb 25, 2006
  4. Feb 25, 2006 #3
    Whoops, just a typo anyway.

    That was just a guesswork based on intuition and my ignorance, I simply don't know. So tell me, why would you expect the latter?
  5. Feb 26, 2006 #4
    Well, at least you could point some links.
  6. Feb 26, 2006 #5


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    Or you could look at the definition of the differential in any calculus book:

    [tex]df(x,y,z)= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz[/tex]
    which, if you go back to the definition of the partial derivatives, is the formula I gave.
  7. Feb 26, 2006 #6
    Whoops, we've got a communication problem here, it seems.
    I know that this is the definition, and I've written it before you as well. My question is: why? What geometry/mathematical quantity does this correntponds to?
  8. Mar 1, 2006 #7
    From what I've found, it should be like this:
    [tex]df(x,y,z)= \frac{\partial f}{\partial x}dx \vec{i} + \frac{\partial f}{\partial y}dy \vec{j} + \frac{\partial f}{\partial z}dz \vec{k} [/tex]
    But this way, for the magnitude of this df, we gotta multiply (dot product) df with itself and square root...
  9. Mar 1, 2006 #8
    Just picture f as a function of two variables instead of three; z = f(x,y). Thus f can be represented by a surface S in 3 space. The total differential df in the equation
    [tex]df(x,y)= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy[/tex]
    would represent the change in the z component of the tangent plane T of f at a point P at coordinates (x,y,z) as we go to some other point P' at (x+delta x,y+delta y,f(x+delta x,y+delta y)). In this case, since delta x and delta y are independent variables, we can say dx=delta x and dy=delta y; i.e dx and dy can be actual distance values. Notice that df is related to the z component of points in T but delta f is related to the z component of points in the surface that f generates.

    It's even easier to picture if you just consider f as function of one variable: y=f(x). Though you lose the "partial" nature of the derivitive, the geometric picture you were looking for is still there. For example, you can picture some curved line that f generates and some tangent line T at a point P at (x, f(x)). If you take another point on the curve P' at (x+delta x, f(x+delta x)), dy would represent the verticle distance between a horizontal line through P and the tangent line T. In other words, dy would represent the distance that T changes in the y direction for some change dx=delta x in the x direction. We could imagine an animal called delta y (as distinct from dy) that represents the distance from the horizontal line through point P to the CURVE (rather than the tangent line T).

    Why would I even begin to set here and type all that??? :surprised
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