Answer
$\triangle z=-0.7189$ and $dz=-0.73$
Work Step by Step
We are given that $z=x^2-xy+3y^2$
The differential form for the given function is:
$dz=(\dfrac{\partial z}{\partial x}) dx + (\dfrac{\partial z}{\partial y}) dy + (\dfrac{\partial z}{\partial z}) dz$
The partial derivatives for the given function with respect to $x$ and $y$ are:
$\dfrac{\partial z}{\partial x}=2x-y$ and $\dfrac{\partial z}{\partial y}=6y-x$
and $\dfrac{\partial z}{\partial x}(3,-1)=7$ and $\dfrac{\partial z}{\partial y}(3,-1)=-9$
Here, $\triangle x=2.96-3=-0.04$ and $\triangle y=-0.95-(-1)=-0.95+1=0.05$
and $dz=(7) (-0.04) +(-9) (0.05)=-0.73$
Hence,
$\triangle z=f(3,-1)-f(2.96,-0.95)=14.2811-15=-0.7189$
