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: 11

Answer

$ 6x+4y-23$

Work Step by Step

Given that f is a differentiable function with $f(x,y)=1+xln(xy-5)$ at $(2,3)$ The linearization $L(x,y)$ of function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ $f(x,y)=1+xln(xy-5)$ $f_{x}(a,b)=ln(xy-5)+\frac{xy}{(xy-5)}$ $f_{y}(a,b)=\frac{x^{2}}{(xy-5)}$ At $(2,3)$ $f_{x}(2,3)=ln((2).(3)-5)+\frac{(2).(3)}{(2*3-5)}=ln1+6=6$ $f_{y}(2,3)=\frac{2^{2}}{(6-5)}=4$ The linearization $L(x,y)$ of function at $(2,3)$ is $L(x,y)= f(2,3)+f_{x}(2,3)(x-2)+f_{y}(2,3)(y-3)$ $= f(2,3)+6(x-2)+4(y-3)$ $=1+6(x-2)+4(y-3)$ $=1+6x-12+4y-12$ $=6x+4y-23$ Hence, the linearization $L(x,y)$ of the function at $(2,3)$ is $ 6x+4y-23$
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