## Thomas' Calculus 13th Edition

Published by Pearson

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

#### Answer

$\dfrac{-3}{2\sqrt {13}}$

#### 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: $h_x(1,1)=\dfrac{-y/x^2}{1+\dfrac{y^2}{x^2}}+\sqrt 3 \cdot \dfrac{(y/2)}{\sqrt {1-(xy/2)^2}}=\dfrac{1}{2} \\ h_y (1,1)=\dfrac{1/x}{1+\dfrac{y^2}{x^2}}+\sqrt 3 \cdot \dfrac{(x/2)}{\sqrt {1-(xy/2)^2}}=\dfrac{3}{2}$ Write the gradient equation. $\nabla f (1,-1)= \lt h_x,h_y \gt = \lt \dfrac{1}{2} , \dfrac{3}{2} \gt$ Thus, $v=\dfrac{u}{|u|} = \dfrac{\lt 3,-2 \gt }{\sqrt {(3)^2+(-2)^2}} =\lt \dfrac{3}{\sqrt {13}}, -\dfrac{2}{\sqrt {13}}\gt$ The directional derivative at that direction is given as: $D_u h=\lt \dfrac{1}{2} , \dfrac{3}{2} \gt \times \lt \dfrac{3}{\sqrt {13}}, -\dfrac{2}{\sqrt {13}}\gt =\dfrac{-3}{2\sqrt {13}}$

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