Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.4 - Tangent Planes and Linear Approximation - 14.4 Exercise - Page 934: 18

Answer

$x+y-1$

Work Step by Step

Given $f(x,y)=)=\dfrac{y-1}{x+1}$ Linear approximation : $L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0$ $z_0=\dfrac{0-1}{0+1}=-1; f_x=-\dfrac{y-1}{(x+1)^2} \\f_y=\dfrac{1}{x+1}$ Then, we have $f_x(0,0)=1; f_y(0,0)=0$ Therefore, $L(x,y)=L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0=1(x-0)+1(y-0)+(-1)=x+y-1$
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