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What is derivative of a vector respect to another vector?

  1. Sep 17, 2010 #1
    I am confused. I never seen derivative of a vector respect to another vector. When I go on the web, the article just show divergence, curl, gradient etc. But not derivative of a vector respect to another vector?

    For example what is

    [tex]\frac{d(\vec{x}-\vec{x_0})^2}{d \vec{x}} ?[/tex]

    For [tex]\vec{x_0}[/tex] is a constant vector.

    The book seems to imply:

    [tex]\frac{d[(\vec{x}-\vec{x_0})^2]}{d \vec{x}} = 2(\vec{x}-\vec{x_0}) \frac{d \vec{x}}{d \vec{x}} = 2(\vec{x}-\vec{x_0}) [/tex]

    I guess I don't know how to do a derivative like this. Can anyone help? I have looked through the multiple variable book and nothing like this. The variable is always scalar. The closest I seen is:

    [tex]\int_C \vec{F} \cdot d\vec{r} \;=\; \int_C \vec{F} \cdot \hat{r}dr[/tex]

    But this is not exactly what the book discribed.

    The only one that is remotely close is Directional Derivative which I don't think so.
    Last edited: Sep 18, 2010
  2. jcsd
  3. Sep 18, 2010 #2


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    Wow, I agree that is really confusing notation.
    Probably, they mean
    [tex](\vec x - \vec x_0)^2 = (\vec x - \vec x_0) \cdot (\vec x - \vec x_0)[/tex]
    so the square is actually a scalar.

    Then in components, you could write
    \left( \frac{\mathrm d [(\vec x - \vec x_0)^2] }{ \mathrm d\vec x } \right)_j =
    \frac{\mathrm d [(\vec x - \vec x_0)^2] }{ \mathrm d\vec x_j } =
    2(\vec x - \vec x_0)_j = 2(\vec x_j - (\vec{x_0})_j) )

    If you doubt this, you can write out
    [tex](\vec x - \vec x_0) \cdot (\vec x - \vec x_0) = \left( \sum_{i = 1}^n (\vec x_i)^2 \right) + 2 \left( \sum_{i = 1}^n (\vec x_i) (\vec x_0)_i \right) + \left( \sum_{i = 1}^n ((\vec{x_0})_i)^2 \right)[/tex]
    and use that
    [tex]\frac{\mathrm d}{\mathrm d \vec x_j} \left( \sum_{i = 1}^n (\vec x_i)^2 \right) = 2 \vec x_j[/tex]
  4. Sep 18, 2010 #3

    George Jones

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    What book?
  5. Sep 18, 2010 #4

    D H

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    The derivative of a vector function of a vector [tex]\vec f(\vec x)[/tex] with respect to a vector [tex]\vec x[/tex] is a 1-1 tensor, with the i,j element being

    [tex]\frac{\partial f_i(\vec x)}{x_j}}[/tex]

    However, [itex](\vec x - \vec x_0)^2[/itex] is not a vector function. It is a scalar. You are just calculating the gradient:

    [tex]\nabla f(\vec x) = \sum_j \frac{\partial f(\vec x)}{x_j}}\hat x_j[/tex]

    Note that the gradient looks a lot like a vector. It is better thought of as being a covector.

    So what about the gradient of [itex]f(\vec x) = (\vec x - \vec x_0)^2[/itex]? Expanding this, we get

    [tex]f(\vec x) = (\vec x - \vec x_0)\cdot (\vec x - \vec x_0) = \sum_i (x_i - x_{0,i})^2[/tex]

    Taking the gradient, the jth of the gradient is

    [tex]\left(\nabla f(\vec x)\right)_j = \sum_i 2 (x_i - x_{0,i}) \frac{\partial x_i}{\partial x_j}
    = \sum_i 2 (x_i - x_{0,i})\delta_{ij} = 2(x_j - x_{0,j})[/tex]
  6. Sep 18, 2010 #5
    The book is PDE by Strauss p194 to p195. It is part of the derivation of the Green's Function for sphere. The part is about normal derivative of G. It talked about derivation respect to [itex]\vec{x}[/itex] and some very funcky statement I still don't understand. But the later part just went back to the ordinary definition of normal derivative:

    [tex]\frac{\partial G}{\partial n} = \nabla G \cdot \hat{n}[/tex]

    and derive the equation accordinary as if nothing happened!!!! So it is a non question at this point. Strauss is not a good book in any stretch. I just cannot find any PDE book that cover the Green's Function and the EM book that I ordered is still in shipment!!!


  7. Sep 19, 2010 #6

    George Jones

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    amazon.com's search function lets me look at some, but not all, of the pages in the book. Do you mean the statement
    on page 185?

    Notice that equation (10) on page 185 gives [itex]G[/itex] as a function of both [itex]\bold{x}[/itex] and [itex]\bold{x}_0[/itex], so the quoted statement just means that normal partial derivatives, gradients, divergences, etc., are with respect to the coordinates of [itex]\bold{x}[/itex] and not with respect to the coordinates of [itex]\bold{x}_0[/itex]. The quoted statement does not actually mean "take the derivative with respect to a vector."
  8. Sep 19, 2010 #7
    Yes, that is the sentence I refer to. I just took it literally that derivative with respect to [itex]\vec{x}[/itex]. I have absolutely no issue on the normal derivative. That is the reason I reposted that I have no question on the complete derivation.

    I have not gone into the exercise yet. I still have one more question regarding to a zero vector in my other post that I got stuck!!! If you can help, that would be really appreciated.

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