Answer
$\triangle z=-0.7189$ and $dz=-0.73$
Work Step by Step
Given the function $z=x^2-xy+3y^2$
The differential form can be evaluated as follows:
$dz=\dfrac{\partial z}{\partial x} dx + \dfrac{\partial z}{\partial y} dy + \dfrac{\partial z}{\partial z} dz$
We need to find the partial derivatives w.r.t. $t$ and $x$ as follows:
$z_x=2x-y\\z_y=6y-x$
and $z_x(3,-1)=7\\z_y(3,-1)=-9$
Here, $\triangle x=2.96-3=-0.04$ and $\triangle y=-0.95-(-1)=0.05$
Now, we have
$dz=7 \cdot (-0.04) +(-9) \cdot (0.05)=-0.73$
Thus,
$\triangle z=f(3,-1)-f(2.96,-0.95)=14.2811-15=-0.7189$
Hence, $\triangle z=-0.7189$ and $dz=-0.73$
