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Derivative of a dot product

  1. Jun 17, 2006 #1
    I have come across something that seems a little strange to me. The derivative of a dot product is something similar to the product rule. I am having difficulties grasping this. Isn't the dot product of two vectors a scalar? And then I always thought of a scalar as a real number and the derivative of a real number is 0. So why does a theorem similar to the product rule exist and work?
  2. jcsd
  3. Jun 17, 2006 #2


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    A "scalar" is generally a "scalar function"... which is not necessarily a "scalar constant". The derivative of a constant is what is zero.
  4. Jun 18, 2006 #3
    Of two constant vectors, yes, the dot product is a constant (and a scalar). But when you consider vector functions, e.g.
    T(x)=exp(x)i + log(x)j
    U(x)=cos(x)i + csc(x)j
    Then the dot-product of these will definitely not be a constant -- it will be the quantity exp(x)cos(x) + log(x)csc(x). That's where the formula is useful. You'll find that if T and U are constant vectors, then the formula will give you the expected result: 0.
    Last edited: Jun 18, 2006
  5. Jun 19, 2006 #4


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    product rule

    it helps to know what a derivative is, i.e. that a derivative is a continuous linear map that approximates the difference in the values of the original map.

    Take the multiplication map RxR-->R taking (x,y)-->xy. This is a bilinear map and its derivative at (a,b) is a linear map L(s,t) such that if x = a+s, y = b+t, then the difference xy-ab is approximated by the linear map L(s,t) to within an error which vanishes faster than (s,t) as (x,y) approaches (a,b).

    In this case take L(s,t) = at+bs. Then the difference xy-ab differs from at+bs by the error term xy-ab - at-bs = (a+s)(b+t)-ab-at-bs = st, which does vanish faster than either s or t as (s,t) goes to zero.

    so the derivative of xy, at (a,b), as a linear map on RxR, is the map taking (s,t) to at+bs. To get the leibniz rule, compose the derivative (f'(c),g'(c)) of a differentiable map R->RxR defined by a pair (f,g), with the derivative of multiplication taken at (f(c),g(c)).

    This gives the linear map taking r to (f'(c)g(c)+f(c)g'(c))r.

    as usually taught in elemenmtary calculus, this linear map is referred to only by the unique entry in tis 1by1 matrix, namely the number f'(c)g(c)+f(c)g'(c).

    The same proof works for any continuous bilinear map, modulo the fact that commutativity does not always hold.

    So the derivative of a possibly non commutative continuous bilinear map VxW-->U at a point (a,b), where V,W,U are complete normed vector spaces, maps (s,t) to (bs+at). and so the derivative of a product map fg:X-->U, viewed as a composite X-->VxV-->U where VxV-->U is a bilinear product, and (f,g) define the map X-->VxV, at the point c in X, is the linear map X-->U taking x in X to, what else? some plausible continuous linear function of x with values in U, namely [I guess] f’(c)(x).g(c) + f(c).g’(c)(x).

    Hows that look?
    Last edited: Jun 19, 2006
  6. Jun 19, 2006 #5


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    if this seems confusing, it is. i not long ago had an anonymous referee for a research paper of mine, object to a statement i made about the derivative of a bilinear map, which statement however was completely correct. so even professionals are confused by these things at times.

    but this is a lesson in elarning the right definition of a derivative, as a linear approximation to the difference map f(x)-f(a), linear in (x-a) that is.

    i also had a professional analyst friend make a mistake on this latter score in a prelim exam, confusing linearity in x with linearity in (x-a).
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