Chapter 9 - Multiveriable Calculus - 9.5 Total Differentials and Approximations - 9.5 Exercises - Page 502: 2

dz=0.49

We are given $z=f(x,y)=5x^{3}+2xy^{2}-4y$ $f_{x}(x,y)=15x^{2}+2y^{2}$ $f_{y}(x,y)=4xy-4$ By the definition $dz=f_{x}(x,y).dx+f_{y}(x,y)dy$ $dz=(15x^{2}+2y^{2})dx+(4xy-4)dy$ $dz=[15\times(1)^{2}+2\times(3)^{2}]\times0.01+[4\times1\times3-4]\times0.02=0.49$