Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - Review - Exercises - Page 1189: 22


$\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$

Work Step by Step

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 $\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$
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