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: 8


$\lt -\dfrac{11}{2},-6, \dfrac{1}{2} \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: $f_x=-6x+\dfrac{z}{1+(xz)^2} =\dfrac{z}{1+x^2z^2}-6x\\f_y=-6yz \\f_z=6z^2-3x^2-3y^2+\dfrac{x}{1+x^2z^2}$ Write the gradient vector equation. $\nabla f = \lt f_x,f_y,f_z \gt $ For point $(1,1,1)$ $\nabla f = \lt \dfrac{z}{1+x^2z^2}-6x, -6yz, 6z^2-3x^2-3y^2+\dfrac{x}{1+x^2z^2} \gt = \lt -\dfrac{11}{2},-6, \dfrac{1}{2} \gt $
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