Identity for $\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d}$ I am looking for an identity for the following derivative:
$$\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d},$$
where $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$ are column vectors.
My approach to expand the expression is as follows:
\begin{align*}
\nabla[( \mathbf{a} \cdot \mathbf{b} )\mathbf{c}] \cdot \mathbf{d}
&= [ ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} + \mathbf{c} \otimes\nabla(\mathbf{a} \cdot \mathbf{b}) ] \cdot \mathbf{d} \\
&= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ \mathbf{c} \otimes [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \} \cdot \mathbf{d}\\
&= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ \mathbf{c} \otimes [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \}^\top \mathbf{d}\\
&= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + \{ [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}] \otimes \mathbf{c} \} \mathbf{d}\\
&= ( \mathbf{a} \cdot \mathbf{b} ) D \mathbf{c} \cdot \mathbf{d} + (\mathbf{c} \cdot \mathbf{d}) [ (D\mathbf{a})^\top \mathbf{b} + (D\mathbf{b})^\top \mathbf{a}].
\end{align*}
Here, $D \equiv \nabla^\top$, i.e., $\nabla$ stands for the gradient operator and $D$ denotes the Jacobian.
My question is, is this expansion correct? Moreover, in my understanding, the resulting expression is again a column vector. Is it right?
 A: It is said in comments that OP wants to compute the directional derivative in direction $d$. I'll write $v$ instead of $d$. The usual extension of a map defined for scalar-fields to vector fields is to apply the known map componentwise, at least in cartesian coordinates (see e.g. the vector Laplacian). That is, I would write
$${\let\del\partial} \big((v\cdot \nabla) ((a\cdot b) c)\big)_k: = v_i\del_i ( a_j b_j c_k)$$
one has by the usual 1D Leibniz rule
$$ v_i\del_i ( a_j b_j c_k) =  (v_i\del_ia_j) b_j c_k +  a_j (v_i\del_ib_j) c_k +  a_j b_j v_i\del_ic_k$$
which can be also written
$$ (v\cdot \nabla) ((a\cdot b) c) =  (((v\cdot \nabla)a)\cdot b) c+(a\cdot ((v\cdot \nabla)b)) c+(a\cdot b) (v\cdot \nabla)c$$
I would write $(\nabla f)_{ij} = \del_j f_i$ and hence $ v_i\del_i f_j = (\nabla f^Tv)_j $. So you can rewrite the above only using matrix products, if you wanted:
$$((v\cdot \nabla)a)\cdot b = ((v\cdot \nabla)a)^Tb=(\nabla a^Tv)^Tb = v^T(\nabla a)b ,
\\
a\cdot ((v\cdot \nabla)b) =a^T\nabla b^Tv 
\\ (a\cdot b) (v\cdot \nabla)c = a^Tb \nabla c^T v$$
i.e.
$$ (v\cdot \nabla) ((a\cdot b) c) = (v^T(\nabla a)b)c + (a^T\nabla b^Tv)c+ a^Tb \nabla c^T v$$
A: I think I arrive at the same conclusion but in slightly different way.
Using differentials
$$
d\mathbf{f} = (d\mathbf{a})^T\mathbf{b} \mathbf{c}+
\mathbf{a}^T d\mathbf{b} \mathbf{c}+
\mathbf{a}^T\mathbf{b} d\mathbf{c}
$$
Now using
$d\mathbf{a}=\mathbf{J}_a d\mathbf{x}$... and rearranging terms,
you will find that the Jacobian of $\mathbf{f}$ which is a matrix, is
$$
\mathbf{J}_f = 
\mathbf{c} 
\left[
\mathbf{b}^T \mathbf{J}_a + \mathbf{a}^T \mathbf{J}_b 
\right] +
\mathbf{a}^T\mathbf{b} \mathbf{J}_c 
$$
The directional derivative is $\mathbf{J}_f \mathbf{d}$.
