Showing that $ (v \cdot \nabla) v = (\nabla \times v) \times v + \frac{1}{2} \nabla (v \cdot v)$ (Fluid Mechanics , Feynman) Right under equation 40.7 in the feynman lecture (here), this identity is written:
$$ (v \cdot \nabla) v = (\nabla \times v) \times v + \frac{1}{2} \nabla (v \cdot v)$$
I seek a proof for this identity/ an intuitive proof for why it is true. I'm not sure how I'd even start the derivation but I think this identity is the same as the one under the 'special sections' part of this wiki page
An attempt:
By the vector triple product identity
$$ a \times b \times c = (b ) c \cdot a - ( c ) b \cdot a$$
Now this gave me zero when applied to $ \nabla \times v \times v$ and that doesn't look right..
Hints would also be appreciated :)
 A: In the equation you have, in the $(\nabla \times \ \textbf{v}) \times \textbf{v}$ term, there are parantheses around the first two terms, so it's not necessarily zero since first you do $\nabla \times \ \textbf{v}$, which is perpendicular to $\textbf{v}$, and then you cross it with $\textbf{v}$.
Starting with the vector triple product seems like a good approach. You can always write it out in tensor notation with the Levi-Civita symbol to get the $n^{th}$ component:
$$\left((\partial_i v_j) \epsilon_{ijk}\right) v_m \epsilon_{kmn},$$
where $\left((\partial_i v_j) \epsilon_{ijk}\right)$ is the $k^{th}$ component of $(\nabla \times \ \textbf{v})$.
Since $\epsilon_{kmn} = \epsilon_{mnk}$ and $\epsilon_{ijk}\epsilon_{mnk}=\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$, the $n^{th}$ component is equal to
$$ \begin{eqnarray}(\partial_i v_j)  v_m \epsilon_{ijk}\epsilon_{mnk}\\
 &=&(\partial_i v_j) v_m (\delta_{im}\delta_{jn}-\delta_{in}\delta_{jm})\\
&=&(\partial_i v_j) v_m \delta_{im}\delta_{jn}-(\partial_i v_j) v_m\delta_{in}\delta_{jm}\\
&=& v_m \partial_m   v_n -\partial_n v_m v_m.
\end{eqnarray}$$
Now, you can use what @PAM1499 noted (the chain rule) to write $\partial_n v_m v_m = \frac{1}{2} \partial_n (v_m v_m)$.
Putting everything back into vector form, you get:
$$(\nabla \times \ \textbf{v}) \times \textbf{v} = (\textbf{v} \cdot \nabla ) \textbf{v}- \frac{1}{2} \nabla(\textbf{v} \cdot \textbf{v}).$$
and all you need to do is rearrange the terms to get the identity.
