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: 13



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: $g_x=\dfrac{(xy+2)(1)-(x-y)(y)}{(xy+2)^2}=\dfrac{y^2+2}{(xy+2)^2} \\ g_y=\dfrac{(xy+2)(-1)-(x-y)(x)}{(xy+2)^2}=-\dfrac{x^2+2}{(xy+2)^2} $ Write the gradient equation. $\nabla f (1,-1)= \lt f_x,f_y \gt = \lt \dfrac{(-1)^2+2}{(-1+2)^2}, -\dfrac{1+2}{(-1+2)^2} \gt =\lt 3,-3 \gt$ Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 12, 5 \gt }{\sqrt {(12)^2+(5)^2}} =\lt \dfrac{12}{13}, \dfrac{5}{13} \gt$ The directional derivative at that direction is given as: $D_v f=\nabla f \cdot v=\lt 3,-3 \gt \times\lt \dfrac{12}{13}, \dfrac{5}{13} \gt =\dfrac{21}{13}$
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