Answer
Thus, $(1.92^{2}+2.1^{2})^{1/3} = 2+ 0.0067= 2.0067$
Calculator gives $ (1.92^{2}+2.1^{2})^{1/3}$$\approx2.008$
The error is approximately 0.0013
Work Step by Step
We are given $(1.92^{2}+2.1^{2})^{1/3}$
We know that $(2^{2}+2^{2})^{1/3}=2$. Therefore, let $x=2, dx=-0.08,y=2,dy=0.1$
$dz=f_{x}(x,y).dx+f_{y}(x,y).dy$
$dz=(\frac{(x^{2}+y^{2})^{-2/3}.2x}{3})dx+(\frac{(x^{2}+y^{2})^{-2/3}.2y}{3})dy$
$dz=\frac{1}{3}dx+\frac{1}{3}dy$
$dz\approx0.0067$
Thus, $(1.92^{2}+2.1^{2})^{1/3} = 2+ 0.0067= 2.0067$
Calculator gives $ (1.92^{2}+2.1^{2})^{1/3} \approx2.008$
The error is approximately 0.0013
