Gradient of cross product of two vectors (where first is constant) In some book about continuum mechanics I read that from principle of virtual work follows balance of rotational momentum when $\delta \boldsymbol{r} = \boldsymbol{\delta \varphi} \times \boldsymbol{r}, \; \boldsymbol{\delta \varphi} = \boldsymbol{\mathsf{const}}$ ($\boldsymbol{r}$ is location vector, $\delta \boldsymbol{r}$ is its variation, $\boldsymbol{\delta \varphi}$ is not variation, just denoted as it for some reason like being small enough for infinitesimal $\delta \boldsymbol{r}$). Then there is written without any explaination $\boldsymbol{\nabla} \delta \boldsymbol{r} = - \boldsymbol{E} \times \boldsymbol{\delta \varphi}$. I know that $\boldsymbol{E}$ is bivalent “metric unit identity” tensor (the one which is neutral to dot product operation), and that $\boldsymbol{\nabla} \boldsymbol{r} = \boldsymbol{E}$. And that $\boldsymbol{a} \times \boldsymbol{E} = \boldsymbol{E} \times \boldsymbol{a} \:\: \forall\boldsymbol{a}$, no minus here. To get minus, transposing is needed: $\left( \boldsymbol{E} \times \boldsymbol{\delta \varphi} \right)^{\mathsf{T}} \! = - \boldsymbol{E} \times \boldsymbol{\delta \varphi}$. Thus I can’t get why $\boldsymbol{\nabla} \delta \boldsymbol{r} = - \boldsymbol{E} \times \boldsymbol{\delta \varphi}$ has minus sign.
For constant $\boldsymbol{\delta \varphi}$, $\boldsymbol{\nabla} \boldsymbol{\delta \varphi} = {^2\boldsymbol{0}}$ (bivalent zero tensor). Isn’t it true that $\boldsymbol{\nabla} \! \left( \boldsymbol{\delta \varphi} \times \boldsymbol{r} \right) = \boldsymbol{\delta \varphi} \times \boldsymbol{\nabla} \boldsymbol{r} = \boldsymbol{\delta \varphi} \times \boldsymbol{E} = \boldsymbol{E} \times \boldsymbol{\delta \varphi}$? Searching for how to get gradient of cross product of two vectors gives gradient of dot product, divergence ($\boldsymbol{\nabla} \cdot$) of cross product, and many other relations. But no gradient of cross product $\boldsymbol{\nabla} \! \left( \boldsymbol{a} \times \boldsymbol{b} \right) = \ldots$ Is it impossible or unknown how to find it? At least for the case when first vector is constant.
update
As “gradient” I mean tensor product with “nabla” $\boldsymbol{\nabla}$: $\operatorname{^{+1}grad} \boldsymbol{A} \equiv \boldsymbol{\nabla} \! \boldsymbol{A}$, here $\boldsymbol{A}$ may be tensor of any valence (and I don’t use “$\otimes$” or any other symbol for tensor product). Nabla (differential Hamilton’s operator) is $\boldsymbol{\nabla} \equiv (\sum_i)\, \boldsymbol{r}^i \partial_i$, $\:(\sum_i)\, \boldsymbol{r}^i \boldsymbol{r}_i = \boldsymbol{E} \,\Leftrightarrow\, \boldsymbol{r}^i \cdot \boldsymbol{r}_j = \delta^{i}_{j}$ (Kronecker’s delta), $\,\boldsymbol{r}_i \equiv \partial_i \boldsymbol{r}$ (basis vectors), $\,\partial_i \equiv \frac{\partial}{\partial q^i}$, $\:\boldsymbol{r}(q^i)$ is location vector, and $q^i$ $(i = 1, 2, 3)$ are coordinates.
 A: Well, it’s easy to find such a gradient. You mentioned almost everything you need for that but the following

*

*the anticommutativity ${\; \boldsymbol{p} \times \boldsymbol{q} = - \, \boldsymbol{q} \times \boldsymbol{p} \;}$ for any two vectors ${\; \boldsymbol{p} \;}$ and ${\; \boldsymbol{q}}$


*a partial derivative of any vector with respect to scalar like a coordinate isn’t some more complex tensor, it is a vector too


*for differentiation of a “$\circ$”-product of two multipliers, the famous “product rule” https://en.wikipedia.org/wiki/Product_rule applies:
\begin{equation*}
   \displaystyle
   \frac{\partial}{\partial q^i} \bigl( u \circ v \bigr)
   =
   \biggl( \frac{\partial}{\partial q^i} \, u \biggr) \! \circ v
      \, + \,
      u \circ \! \biggl( \frac{\partial}{\partial q^i} \, v \biggr)
\end{equation*}
So we have
\begin{gather*}
   {\boldsymbol{\nabla} \bigl( \boldsymbol{a} \times \boldsymbol{b} \bigr)}
   = {\boldsymbol{r}^i \partial_i \!\, \bigl( \boldsymbol{a} \times \boldsymbol{b} \bigr)}
   = {\boldsymbol{r}^i \! \left(
      \partial_i \boldsymbol{a} \times \boldsymbol{b}
      + \boldsymbol{a} \times \partial_i \boldsymbol{b}
   \right)}
   = {\boldsymbol{r}^i \! \left(
      \partial_i \boldsymbol{a} \times \boldsymbol{b}
      - \partial_i \boldsymbol{b} \times \boldsymbol{a}
   \right)}
   = {\boldsymbol{r}^i \partial_i \boldsymbol{a} \times \boldsymbol{b}
      - \boldsymbol{r}^i \partial_i \boldsymbol{b} \times \boldsymbol{a}}
   = {\boldsymbol{\nabla} \boldsymbol{a} \times \boldsymbol{b}
      - \!
      \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{a}}
\end{gather*}
A minus sign before a second addend appears, because we need to swap multipliers to get the differentiation of the first multiplier to get the full $\boldsymbol{\nabla}$
When a first multiplier, say $\boldsymbol{\phi}$, is constant in space, that is it remains the same although coordinates change, then it becomes
$$
{\boldsymbol{\nabla} \bigl( \boldsymbol{\phi} \times \boldsymbol{b} \bigr)}
= {\boldsymbol{\nabla} \!\, \boldsymbol{\phi} \times \boldsymbol{b} -
\! \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{\phi}}
= {{^2{\boldsymbol{0}}} \times \boldsymbol{b} - \! \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{\phi}}
= {{^2{\boldsymbol{0}}} - \! \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{\phi}}
= {- \boldsymbol{\nabla} \boldsymbol{b} \times \boldsymbol{\phi}}
$$
That’s why there’s a minus sign.
