Answer
$\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$
Work Step by Step
Our aim is to prove that $\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$
Here, $f$ and $g$ defines twice differentiable functions.
Apply the rule as: $\nabla(fg)=f \nabla g+g \nabla f $
Also, we know that $\nabla(f+g)=\nabla f+\nabla g $
$\nabla^2(fg)=\nabla (f \nabla g+g \nabla f )=\nabla (f \nabla g)+\nabla (g \nabla f)$
Further, we have
$\nabla (f \nabla g)+\nabla (g \nabla f)=[f \nabla^2 g+(\nabla f)(\nabla g)]+[g \nabla^2 f+(\nabla g)(\nabla f)]$
or, $\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$
Thus, the result has been proved.
