Answer
\begin{aligned} \text { (a) } f(2,1) & =1 \\ f(2.1,1.05) & =1.05 \\ \Delta z &=0.05 \\ \text {( b) }
dz &=0.05 \end{aligned}
Work Step by Step
Given $$f(x, y)=2 x-3 y$$
so, we have
\begin{aligned} \text { (a) } f(2,1) &=2(2)-3(1)=1 \\ f(2.1,1.05) &=2(2.1)-3(1.05)=4.2-3.15=1.05 \\ \Delta z &=f(2.1,1.05)-f(2,1)=1.05-1=0.05 \\ \text {( b) } \text { since }d z &=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy,\\
&\frac{\partial f}{\partial x}= 2,\\
&\frac{\partial f}{\partial y} =-3,\\
& dx\approx\Delta x=2.1-2=0.1 \ \text{and} \\& dy\approx\Delta y=1.05-1=0.05 \\& \text{So we get},\\
&dz=2 d x-3 d y=2(0.1)-3(0.05)=0.05 \end{aligned}
