Answer
\begin{aligned} \text { (a) } f(2,1) & =0.5 \\ f(2.1,1.05) & =0.5&\\ \Delta z&= 0\\ \text {( b) }
dz &=0\end{aligned}
Work Step by Step
Given $$f(x, y)=\frac{y}{x}$$
so, we have
\begin{aligned} \text { (a) } f(2,1) &=\frac{1}{2}=0.5\\ f(2.1,1.05) &=\frac{1.05}{2.1}=0.5\\
\Delta z &=f(2.1,1.05)-f(2,1)\\
&=0.5-0.5=0 \\
\text {( b) } \text { since }d z &=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy,\\
&\frac{\partial f}{\partial x}= -\frac{y}{x^2}=-\frac{1}{2^2}=-0.25,\\
&\frac{\partial f}{\partial y} =\frac{1}{x}=-\frac{1}{2}=0.5,\\
& 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=-0.25 d x+0.5 d y\\
&\ \ \ \ =-0.25(0.1)+0.5(0.05) \\
&\ \ \ \ =-0.025+0.025=0 \end{aligned}
