We know from the tensor calculus that: $\vec\nabla (a\cdot b) = b\vec\nabla a + a \vec\nabla b$ , where $a$ and $b$ are two scalar functions.
But in the case where for example $a$ is a scalar function and $b$ is a vector how to develop that expression of gradient?
$$\vec{\nabla}\left(a\cdot \vec v \right) = ?$$
Best Answer
These sort of identities are usually proved in the component form and then transferred back to component-free form. In view of this, note that $\nabla(a\boldsymbol{v})$ is a second order tensor. Thus using the product rule,
$$\left(\nabla(a\boldsymbol{v})\right)_{ij} = \frac{\partial}{\partial x_j}\left(av_i\right)=\frac{\partial a}{\partial x_j}v_i+a\frac{\partial v_i}{\partial x_j}.$$
From the above component form, it is recognized that
$$\nabla(a\boldsymbol{v}) = \boldsymbol{v}\otimes\nabla a + a\nabla\boldsymbol{v}.$$