expanding tensor notation of Navier stokes equation I'm trying to expand the variable density and viscosity Navier-stokes equation for incompressible flows but I've had no luck so far.
The Navier-Stokes in tensor notation is:
$$
\rho \dfrac{Du_i}{Dt} = -\dfrac{\partial p}{\partial x_i} + \dfrac{\partial}{\partial x_j}\left(\mu \left( \dfrac{\partial u_i}{\partial x_j} + \dfrac{\partial u_j}{\partial x_i} \right) \right)
$$
I'm confused with how to expand the last term in the above equation. The similar term in vector notation is $\nabla.(\mu(\nabla \underline{u}+\nabla\underline{u}^T))$
I've tried to expand the term $\dfrac{\partial}{\partial x_j}\left(\mu \left( \dfrac{\partial u_i}{\partial x_j} + \dfrac{\partial u_j}{\partial x_i} \right) \right)$ in x direction as :
$$
(\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y})(\mu(\dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} + \dfrac{\partial u}{\partial x} + \dfrac{\partial v}{\partial x}))
$$
$$
= \dfrac{\partial}{\partial x}(\mu(2\dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} + \dfrac{\partial v}{\partial x})) + \dfrac{\partial}{\partial y}(\mu(2\dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} + \dfrac{\partial v}{\partial x}))
$$
Is this expansion correct? 
Thank you
 A: Note that
$$\nabla \cdot [\mu (\nabla \mathbf{u}+\nabla \mathbf{u}^T)]=\nabla \mu \cdot(\nabla \mathbf{u}+\nabla \mathbf{u}^T)+ \mu \nabla^2\mathbf{u} + \mu \nabla(\nabla \cdot \mathbf{u}).$$
In terms of Cartesian components, the $\mathbf{e}_i$ component is (summing over $j =1,2$ in $2$D or $j =1,2,3$ in $3$D):
$$\frac{\partial}{\partial x_j}\left[\mu \left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right)\right]=\frac{\partial \mu}{\partial x_j}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right)+\mu \frac{\partial^2 u_i}{\partial x_j \partial x_j}+\mu\frac{\partial}{\partial x_i}\left(\frac{\partial u_j}{\partial x_j}\right).$$
Note that the Einstein summation convention is used here:
$$\frac{\partial^2 u_i}{\partial x_j \partial x_j}=\sum_{j=1}^3\frac{\partial^2 u_i}{\partial x_j^2}=\frac{\partial^2 u_i}{\partial x_1^2}+\frac{\partial^2 u_i}{\partial x_2^2}+\frac{\partial^2 u_i}{\partial x_3^2}$$
In $2$D with $u = u_1$, $v = u_2$, $x = x_1$ and  $y = x_2$, the x-component is
$$\frac{\partial}{\partial x_j}\left[\mu \left(\frac{\partial u_1}{\partial x_j}+\frac{\partial u_j}{\partial x_1} \right)\right]=\frac{\partial}{\partial x}\left(2 \mu  \frac{\partial u}{\partial x} \right)+\frac{\partial}{\partial y}\left[ \mu  \left(\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \right)\right],$$
and the y-component is 
$$\frac{\partial}{\partial x_j}\left[\mu \left(\frac{\partial u_2}{\partial x_j}+\frac{\partial u_j}{\partial x_2} \right)\right]=\frac{\partial}{\partial x}\left[ \mu  \left(\frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \right)\right]+\frac{\partial}{\partial y}\left(2 \mu  \frac{\partial v}{\partial y} \right).$$
