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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 <f,g> ?

May 4th, 2007, 08:03 AM  #3 
Member Joined: Nov 2006 From: Vancouver, Canada Posts: 46 Thanks: 0 
indeed I mean <f,g>. 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 CauchySchwarz inequality), by computing: <x+h1,y+h2>=<x,y>+<x,h2>+<h1,y>+<h1,h2> you realize that: <x+h1,y+h2><x,y>f(h1,h2)=o((h1,h2)), where f(h=(h1,h2))=<x,h2>+<h1,y> is linear and continuous in h, and therefore is the derivative at (x,y) of our scalar product. We have that <h1,h2>=o(h) because <h1,h2> <= h1.h2 by CauchySchwarz. 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><x,x> at point h, then you find Dg(x)(h)=2<x,h> If you are trying to compute the derivative of the norm: N: x>sqrt(<x,x>), then using the Chain rule, you find that DN(x)(h)=<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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