Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.5 Total Differentials and Approximations - 9.5 Exercises - Page 502: 3



Work Step by Step

We are given $z=f(x,y)=\frac{y^{2}+3x}{y^{2}-x}$ $f_{x}(x,y)=\frac{3(y^{2}-x)-(-1)(y^{2}+3x)}{(y^{2}-x)^{2}}$ $f_{x}(x,y)=\frac{3y^{2}-3x+y^{2}+3x}{(y^{2}-x)^{2}}$ $f_{x}(x,y)=\frac{4y^{2}}{(y^{2}-x)^{2}}$ $f_{y}(x,y)=\frac{2y(y^{2}-x)-2y(y^{2}+3x)}{(y^{2}-x)^{2}}$ $f_{}(x,y)=\frac{2y^{3}-2xy-2y^{3}-6xy}{(y^{2}-x)^{2}}$ $f_{}(x,y)=\frac{-8xy}{(y^{2}-x)^{2}}$ By the definition $dz=f_{x}(x,y).dx+f_{y}(x,y)dy$ $dz=(\frac{4y^{2}}{(y^{2}-x)^{2}})dx+(\frac{-8xy}{(y^{2}-x)^{2}})dy$ With $x=4, y=-4, dx=0.01, dy=0.03$ $dz=\frac{7}{225}$
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