Proving vector identities using Cartesian tensor notation

[QuanticEnigma]

Homework Statement

1. Establish the vector identity
$$<br /> \nabla . (B$$ x $$A) = (\nabla$$ x $$A).B - A.(\nabla$$ x $$B)$$

2. Calculate the partial derivative with respect to $$x_{k}$$ of the quadratic form $$A_{rs}x_{r}x_{s}$$ with the $$A_{rs}$$ all constant, i.e. calculate $$A_{rs}x_{r}x_{s,k}$$

Homework Equations

The Attempt at a Solution

1.
$$<br /> \nabla . (B$$ x $$A) = \epsilon_{ijk}A_{j}B_{k,i}$$

Now I don't know what to do next.

2.

$$A_{rs}x_{r}x_{s,k} = A_{rs}\partial_{k}(x_{r}x_{s}) = A_{rs}(x_{r}\partial_k x_{s} + x_{s}\partial_k x_{r})$$

I have no idea if this is right or not.

I'm pretty good at proving vector identities (and Cartesian tensor notation in general), but I get lost when partial derivatives/nablas are involved. Any tips would be greatly appreciated!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[vela]

Your notation is unclear and misleading. When you write $\nabla\cdot(B\times A) = \epsilon_{ijk}A_j B_{k,i}$, it looks like you're only differentiating B_k. Also, you got the indices wrong since you have BxA, not AxB. On the other hand, you should have $\nabla\cdot(A \times B)$ on the LHS anyway.

Instead, you should write

$$\nabla\cdot(A \times B) = \partial_i (\epsilon_{ijk} A_j B_k) = \epsilon_{ijk} \partial_i (A_j B_k)$$

Use the product rule to differentiate and then convert back to vector notation.In the second problem, use the fact that the x's are independent, so $\partial_i x_j = 0$ if $i \ne j$ and $\partial_i x_j = 1$ if $i = j$, i.e. $\partial_i x_j = \delta_{ij}$.