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: 15



Work Step by Step

Given that f is a differentiable function with $f(x,y)=4arctan(xy)$ at $(1,1)$ The linearization $L(x,y)$ of function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ $f(x,y)=4arctan(xy)$ $f_{x}(a,b)=\frac{4y}{(1+x^{2}y^{2})}$ $f_{y}(a,b)=\frac{4x}{(1+x^{2}y^{2})}$ At $(1,1)$ $f(1,1)=4arctan(1)=4.\frac{\pi}{4}=\pi$ $f_{x}(1,1)=\frac{4.1}{(1+1^{2}1^{2})}=2$ $f_{y}(1,1)=\frac{4.1}{(1+1^{2}1^{2})}=2$ The linearization $L(x,y)$ of function at $(1,1)$ is $L(x,y)= f(1,1)+f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)$ $=\pi+2(x-1)+2(y-1)$ $=\pi+2x-2+2y-2$ $=2x+2y+(\pi-4)$ Hence, the linearization $L(x,y)$ of the function at $(1,1)$ is $2x+2y+(\pi-4)$
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