Answer
$dz=\frac{7}{250}=0.028$
Work Step by Step
We are given $z=\frac{x+5y}{x-2y}$
$f_{x}(x,y)=\frac{(x-2y)1-(x+5y)1}{(x-2y)^{2}}=\frac{7y}{(x-2y)^{2}}$
$f_{y}(x,y)=\frac{(x-2y)5-(x+5y)(-2)}{(x-2y)^{2}}=\frac{7x}{(x-2y)^{2}}$
By the definition $dz=f_{x}(x,y).dx+f_{y}(x,y).dy$
with $x=1; y=-2; dx=-0.04; dy=0.02$
we have $dz=(\frac{7y}{(x-2y)^{2}})(-0.04)+(\frac{7x}{(x-2y)^{2}})0.02$
$dz=\frac{7}{250}=0.028$
