Answer
\begin{aligned} \text { (a) } f(2,1) & =7.3891 \\ f(2.1,1.05) & =8.5745 \\ \Delta z&= 1.1854 \\ \text {( b) }
dz &=1.108365\end{aligned}
Work Step by Step
Given $$f(x, y)=y e^x$$
so, we have
\begin{aligned}
&\text { (a) } f(2,1) = e^2 =7.3891 \\ f(2.1,1.05) &=1.05 e^{2.1} =8.5745 \\
\Delta z &=f(2.1,1.05)-f(2,1)\\
&= 8.5745 - 7.3891 =1.1854 \\
\text {( b) } \text { since }d z &=\frac{\partial f}{\partial x} dx+\frac{\partial f}{\partial y} dy,\\
&\frac{\partial f}{\partial x}= y e^x = e^2=7.3891,\\
&\frac{\partial f}{\partial y} = e^x =e^2=7.3891,\\
& 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=e^2 d x+e^2 d y\\
& \ \ \ \ =7.3891 (0.1)+ 7.3891 (0.05) \\
&\ \ \ \ = 0.73891+0.369455\\
& \ \ \ \ =1.108365\end{aligned}
