Are you asking for a proof? The proof is trivial (it follows immediately from the form of the jacobian). Let ##U\subseteq \mathbb{R}^{n}## be open and let ##F:U\rightarrow \mathbb{R}^{m}## be differentiable at ##a\in U##. Then the matrix representation of the total derivative of ##F## at ##a##, in the standard basis ##S##, is given by ##(DF(a))_{S} = (\frac{\partial F^j}{\partial x^i}(a))##. We call this matrix representation the jacobian of such a map. Hence ##Tr(DF(a))_{S} = \sum\frac{\partial F^{i}}{\partial x^{i}}(a)##. Thus if ##X:U\subseteq \mathbb{R}^{3}\rightarrow \mathbb{R}^{3}## is a vector field and ##a\in U##, ##Tr(DX)_{S}(a) = \frac{\partial X^{1}}{\partial x^1}(a) + \frac{\partial X^{2}}{\partial x^2}(a) + \frac{\partial X^{3}}{\partial x^3}(a) = (\nabla\cdot X)(a)##.