Answer
To approximate $$0.98 e^{-0.04}.$$
Notice that $0.98\approx 1$ and $-0.04 \approx 0$ and we know that $0.98 e^{-0.04} \approx 0.9416$.
We, therefore, let
$$z=f(x, y)=x e^{y}$$
Then
$$
\begin{aligned}
d z &=f_{x}(x, y) d x+f_{y}(x, y) d y \\
&=e^{y} d x+x e^{y} d y
\end{aligned}
$$
To approximate $0.98 e^{-0.04},$ we let $x=1$ $d x=-0.02, y=0,$ and $d y=-0.04$, so we have:
$$
\begin{aligned}
d z&=e^{0}(-0.02) +1 \cdot e^{0}(-0.04) \\
&=-0.06 \\
f(0.98,-0.04) &=f(1,0)+\Delta z \\
& \approx f(1,0)+d z \\
&=1 \cdot e^{0}-0.06 \\
&=0.94
\end{aligned}
$$
Thus, $0.98 e^{-0.04} \approx 0.94$
Using a calculator $ 0.98 e^{-0.04} \approx 0.9416$.
The error is approximately $0.0016.$
Work Step by Step
To approximate $$0.98 e^{-0.04}.$$
Notice that $0.98\approx 1$ and $-0.04 \approx 0$ and we know that $0.98 e^{-0.04} \approx 0.9416$.
We, therefore, let
$$z=f(x, y)=x e^{y}$$
Then
$$
\begin{aligned}
d z &=f_{x}(x, y) d x+f_{y}(x, y) d y \\
&=e^{y} d x+x e^{y} d y
\end{aligned}
$$
To approximate $0.98 e^{-0.04},$ we let $x=1$ $d x=-0.02, y=0,$ and $d y=-0.04$, so we have:
$$
\begin{aligned}
d z&=e^{0}(-0.02) +1 \cdot e^{0}(-0.04) \\
&=-0.06 \\
f(0.98,-0.04) &=f(1,0)+\Delta z \\
& \approx f(1,0)+d z \\
&=1 \cdot e^{0}-0.06 \\
&=0.94
\end{aligned}
$$
Thus, $0.98 e^{-0.04} \approx 0.94$
Using a calculator $ 0.98 e^{-0.04} \approx 0.9416$.
The error is approximately $0.0016.$
