Answer
$div (F+G)=div F+div G$
Work Step by Step
When $F=ai+bj$, then we have
$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}$
Plug $F=ai+bj; G=ci+dj$
Now, we have $div (F+G)=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial x}+\dfrac{\partial d}{\partial y}$
$\implies div (F+G)=div F+div G$
Hence, $div (F+G)=div (F)+div (G)$ (proved)
