## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Practice Exercises - Page 865: 42

#### Answer

$\sqrt 3$

#### Work Step by Step

Since, $D_u f = \nabla f \cdot u$ We have: $f(x,y,z)=xyz$ $\nabla f =yz i +xz j +xy k$ or,$\nabla f(1,1,1) =(1)\times (1) i +(1) \times (1) j +(1) \times (1) k=i+j+k$ or, $D_u f (1,1,1)=| \nabla f|_{max}=\sqrt {1^2+1^2+1^2}=\sqrt 3$

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