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The identity in question is

$\nabla (\vec{A} \cdot \vec{B}) = \vec{A} \times (\nabla \times\vec{B}) + \vec{B} \times (\nabla \times \vec{A}) + (A \cdot \nabla)\vec{B} + (\vec{B} \cdot \nabla) \vec{A}$

How do we prove that this is true?

which I found in a physics textbook (Griffiths, Electrodynamics).

My attempt is:

$\vec{A} = A_1\hat{i} + A_2\hat{j}$

$\vec{B} = B_1\hat{i} + B_2\hat{j}$

$\nabla (\vec{A}\cdot \vec{B}) = \nabla (A_1B_1 + A_2B_2)$

$=\nabla A_1 B_1 + A_1 \nabla B_1 + \nabla A_2 B_2 + A_2 \nabla B_2$

I would expect to see negative signs somewhere to be able to identify the result of some cross product. Terms such as $\nabla A_1$ are vectors, but I see no negative signs anywhere.

xoux
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    I'm fairly certain that this has been addressed somewhere on this site before, but I am unable to locate a suitable duplicate. Let me just offer the following in the comments here instead. Firstly, assume that $\vec{A}$ and $\vec{B}$ have three components, not two. Secondly, you've performed one set of simplifications on the LHS. Can you do something similar for the RHS? That is, write down $\nabla \times \vec{B}$ in terms of the derivatives of the components of $\vec{B}$, and then compute $\vec{A} \times (\nabla \times \vec{B})$. Similarly, for $(\vec{A} \cdot \nabla) \vec{B}$, etc. ... – The Amplitwist Mar 19 '22 at 12:55
  • Compare what you get with the component-wise description of the LHS that you derived. It's messy, but it will surely get you what you want. Here is a related question, but I'm not sure whether it will fully help you: Vector Calculus Identities Using Differential Forms... – The Amplitwist Mar 19 '22 at 12:56
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    There is a problem in Griffiths which says, verbatim: "(For masochists only) Prove product rules ii) and iv)". ii) happens to be the identity that I am asking about in my question above. So yes, messy. – xoux Mar 19 '22 at 21:25
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    I think there are only two ways out here :) (A) Do this messy calculation precisely once in your lifetime, and then gift yourself a lifetime-valid guilt-free pass to apply it wherever you like; (B) Learn enough advanced math to watch it fall out as a simple corollary to a general result. (Of course, this comment is made tongue-in-cheek. Hope that's alright...) – The Amplitwist Mar 19 '22 at 21:31
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    For the record, in Griffiths' Electrodynamics has solutions manual. The solution to my question is that of Chapter 1, problem 1.23 in the solution manual to the 4th edition. – xoux Mar 20 '22 at 08:56

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