Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.3 Exercises - Page 896: 52

Answer

$${h_x}\left( { - 2,1} \right) = - 4{\text{ and }}{h_y}\left( { - 2,1} \right) = - 2$$

Work Step by Step

$$\eqalign{ & h\left( {x,y} \right) = {x^2} - {y^2},{\text{ }}\left( { - 2,1,3} \right) \cr & {\text{Calculate the partial derivatives }}{h_x}\left( {x,y} \right){\text{ and }}{h_y}\left( {x,y} \right) \cr & {h_x}\left( {x,y} \right) = \frac{d}{{dx}}\left[ {{x^2} - {y^2}} \right] \cr & {h_x}\left( {x,y} \right) = 2x \cr & {\text{Evaluate at }}\left( { - 2,1,3} \right) \cr & {h_x}\left( { - 2,1} \right) = 2\left( { - 2} \right) \cr & {h_x}\left( { - 2,1} \right) = - 4 \cr & and \cr & {h_y}\left( {x,y} \right) = \frac{d}{{dy}}\left[ {{x^2} - {y^2}} \right] \cr & {h_y}\left( {x,y} \right) = - 2y \cr & {\text{Evaluate at }}\left( { - 2,1,3} \right) \cr & {h_y}\left( { - 2,1} \right) = - 2\left( 1 \right) \cr & {h_y}\left( { - 2,1} \right) = - 2 \cr} $$
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