Are the diffusion terms conservative? Generally the diffusion terms are of the form $$D =   \dfrac{\partial}{\partial x} \left(\mu \dfrac{\partial u}{\partial x} \right)  .$$ 
Is this this term conservative or nonconservative?
 A: In terms of fluid dynamics, a conservation law is one in which the net flux in is equal to the net flux out. This is typically represented as the PDE,1
$$
\frac{\partial u}{\partial t}+\nabla\cdot\mathbf F=S\tag{1}
$$
where $u$ is the conserved quantity, $\mathbf F$ the flux and $S$ the source term.2 In your case, $\mathbf F=-\mu\nabla u$, so it is a conservative term because it satisfies (1).
Note, though, that the domain of dependence for a diffusion equation at a point  $\left(\mathbf x,\,t\right)$ is the entire domain at all previous times. This differs from the convective equation where the domain of dependence is along characteristics (lines that satisfy $du/dt=0$).

1. This can be equivalently written as an integral equation.
2. Often times $S=0$.
A: This form is conservative in the sense that, if you approximate the right hand side with a central finite difference approximation (using $\mu$ at the boundary of each grid cell and u at the center of each cell), the finite difference approximation will automatically conserve mass.
For those of us who solve diffusive problems using numerical methods, this is what a conservative form of the diffusion terms represents. An example of the non-conservative form would be if we differentiated by the product rule to obtain the mathematically equivalent form:  $$D=\mu\frac{\partial ^2u}{\partial t^2}+\frac{\partial \mu}{\partial x}\frac{\partial u}{\partial x}$$If this were expressed in finite difference form, the finite difference scheme would not automatically conserve mass. Such a version would be regarded as non-conservative.
