## Calculus: Early Transcendentals 8th Edition

$\triangle z=0.9225$ and $dz=0.9$
We are given that $z=5x^2+y^2$ The differential form of the given function can be evaluated as follows: $dz=10x dx+2y dy$ Here, $\triangle x=1.05-1=0.05$ and $\triangle y=2.1-2=0.1$ Now, we have $dz=10 \cdot (1) (0.05)+2 \cdot (2) (0.1)$ or, $dz=0.9$ Thus, $\triangle z=f(1.05,2.1)-f(1, 2)$ or, $=5(1.05)^2+(2.1)^2-(5+4)$ Thus, $\triangle z=0.9225$