Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.5 - Directional Derivatives and Gradient Vectors - Exercises 14.5 - Page 826: 7


$\lt 3,2,-4 \gt $

Work Step by Step

In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant to find the x-coordinate of the gradient vector, and vice versa: $f_x=2x+\dfrac{z}{x} \\f_y=2y \\f_z= -4z+\ln x$ and $f_y(2,1)=\dfrac{\partial }{\partial x}(y-x)=y^{0}=1|_{(2,1)}$ Write the gradient vector equation. $\nabla f = \lt f_x,f_y,f_z \gt $ For point $(1,1,1)$ $\nabla f = \lt 2x+\dfrac{z}{x},2y , -4z+\ln x\gt = \lt 2(1)+\dfrac{1}{1},2(1) , -4(1)+\ln (1) \gt = \lt 3,2,-4 \gt $
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