## Calculus 8th Edition

$\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$
Let us consider $\nabla(fg)=f \nabla g+g \nabla f$; $\nabla(f+g)=\nabla f+\nabla g$ This implies that $\nabla^2(fg)=\nabla (f \nabla g+g \nabla f )=\nabla (f \nabla g)+\nabla (g \nabla f)$ or, $\nabla (f \nabla g)+\nabla (g \nabla f)=[f \nabla^2 g+(\nabla f)(\nabla g)]+[g \nabla^2 f+(\nabla g)(\nabla f)]$ Thus, we have $\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$