Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Review - Exercises - Page 1149: 22

Answer

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

Work Step by Step

We need to show that $\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$ Here, $f$ and $g$ are twice differentiable functions. We need to use $\nabla(fg)=f \nabla g+g \nabla f $ and $\nabla(f+g)=\nabla f+\nabla g $ Thus, we have$\nabla^2(fg)=\nabla (f \nabla g+g \nabla f )=\nabla (f \nabla g)+\nabla (g \nabla f)$ This implies that $\nabla (f \nabla g)+\nabla (g \nabla f)=[f \nabla^2 g+(\nabla f)(\nabla g)]+[g \nabla^2 f+(\nabla g)(\nabla f)]$ or, $\nabla^2(fg)=f \nabla^2 g+g \nabla^2 f+2 \nabla f\cdot \nabla g$ Hence, the result has been proved.
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