How to prove this with Einstein summation index notation?

Question

Show that

$$\nabla \left ( \mathbf{u} \cdot \mathbf{r} \right ) = \mathbf{u} + \nabla \mathbf{u} \cdot \mathbf{r}$$

where u is any vector field and r is the position vector.

Answer 1

The $i$th component of the left-hand side is$$\partial_i(u_jr_j)=u_j\delta_{ij}+(\partial_iu_j)r_j=u_i+(\partial_iu_j)r_j.$$In this final expression, the first (second) term is the $i$th component of $\vec{u}$ ($\nabla\vec{u}\cdot\vec{r}$).

Answer 2

hint

The $j^{\text{th}} $ component will be

$$\frac{\partial u_ir_i}{\partial x_j}=$$

$$u_i\frac{\partial x_i}{\partial x_j} + x_j\frac{\partial u_i}{\partial x_j} =$$