# Where can I find this formula?

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Hi all: I just met a equation like following: (r|n') = (r|n)' - (r'|n) where r, n are vectors. (A|B) means the inner product, and A' is the derivative of A. Where can I find the corresponding formula of the above equation please? Thanks

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I am not sure I understand the question. Are you asking how to prove that formula? I suggest you try to prove this one instead (which is obviously equivalent):
(r|n)' = (r'|n) + (r|n')

Just write r as (r_1(x), r_2(x), ..., r_k(x)) and n as (n_1(x), n_2(n), ..., n_k(x)) and expand the inner products using the definition. With that and the chain rule, you should be able to complete the proof.

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Could you give me some more details please? I am not very good at vector calculus. Thanks

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It really is just following the definitions. Give it a try. Start with (r|n)' and expand the definition of the inner product. Then use the fact that the derivative of a sum is the sum of the derivatives.

If you don't see the end, post how far you got.

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(r|n)' = (r_1(x)*n_1(x)+r_2(x)*n_2(x)+...r_m(x)*n_m(x))'
= r_1(x) * dn_1(x) / dx + n_1(x)* dr_1(x)/dx + ... + r_m(x) * dn_m(x) / dx + n_m(x)* dr_m(x)/dx

(r'|n) + (r|n') = n_1(x) * dr_1(x)/dx + ...n_m(x) * dr_m(x)/dx + r_1(x) * dn_1 (x)/dx + ... +r_m(x) * dn_m(x) / dx

Therefore (r|n)' == (r'|n) + (r|n')

Am I right???

Thanks

Yep, you got it!

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Rutin
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