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 April 11th, 2007, 09:01 PM #1 Member   Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 Inner product product rule Hi, how would one deduce the derivative product rule for the general inner product of functions? I was told to use chain rule, but I don't see how I could do that.
 April 11th, 2007, 11:45 PM #2 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Do you mean you want to differentiate fg , or the scalar product of f and g, namely ?
 May 4th, 2007, 08:03 AM #3 Member   Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 indeed I mean . Wow haven't been back for awhile!
 May 4th, 2007, 07:47 PM #4 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Given a normed real vector space E and a scalar product <.,.>: E x E -> R (which is continuous for the euclidean norm by Cauchy-Schwarz inequality), by computing: =+++ you realize that: --f(h1,h2)=o((h1,h2)), where f(h=(h1,h2))=+ is linear and continuous in h, and therefore is the derivative at (x,y) of our scalar product. We have that =o(h) because || <= ||h1||.||h2|| by Cauchy-Schwarz. This is a standard result for all continuous bilinear applications, and can be generalized to multinilear applications by induction. If you are trying to compute the derivative of g: x-> at point h, then you find Dg(x)(h)=2 If you are trying to compute the derivative of the norm: N: x->sqrt(), then using the Chain rule, you find that DN(x)(h)=/N(x).
 May 4th, 2007, 11:26 PM #5 Member   Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 Oh I see... so you mean f(h) is the derivative operator, so then f(h)(x) is the derivative at x?
 May 5th, 2007, 12:31 AM #6 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 I should have typed f_(x,y)(h1,h2). The derivative of f at point (x,y) (evaluated at (h1,h2)). Usually the easiest way to evaluate the derivative of a function say g at point x is to compute limit( [g(x+a*h)-g(x)]/a, a->0 ). This gives you the candidate Dg(x)(h). You then have to verify that Dg(x) is indeed linear continuous (the continuity always holds in finite dimension) and that g(x+h)-g(x)-Dg(x)(h)=o(h) in order to conclude that Dg(x) is the derivative of g at point x. This method happens to be really useful for complicated expressions, for instance calculating the derivative of the Determinant. For instance, what is the derivative of the application: Det: Mat(R^n)->R given by Det(M)= determinant of M for every matrix M ?
 May 6th, 2007, 04:27 PM #7 Member   Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 So... the Det: operation is a function... and you mean the derivative is taken with respect to these functions? I guess I could use chain rule??? Ummm.. (looking at the ceiling lol).

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