Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises - Page 974: 17


$e^xcos(xy) \approx x+1$

Work Step by Step

The linear approximation is provided by the formula: $L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0$ Given $f(x,y)=e^xcos(xy)$ we find that: $z_0=f(0,0)=e^0cos(0*0)=1$ $f_x=e^xcos(xy)-ye^xsin(x)$ $f_y=-xe^xsin(xy)$ Evaluating the functions at the point $(0,0)$ $f_x(0,0)=1$ $f_y(0,0)=0$ Using the linear approximation formula, the linear approximation of $f$ at $(x_0,y_0)=(0,0)$ becomes: $L(x,y)=1(x-0)+0(y-0)+1=x+1$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.