Answer
Using the linear approximation: $\frac{{7.98}}{{2.02}} \simeq 3.95$
Using a calculator: $\frac{{7.98}}{{2.02}} \simeq 3.9505$
Work Step by Step
We are given $f\left( {x,y} \right) = x{\left( {1 + y} \right)^{ - 1}}$. The partial derivatives are
${f_x} = {\left( {1 + y} \right)^{ - 1}}$, ${\ \ \ }$ ${f_y} = - x{\left( {1 + y} \right)^{ - 2}}$
According to Eq. (3), the linear approximation to $f\left( {x,y} \right)$ is given by
$f\left( {a + h,b + k} \right) \approx f\left( {a,b} \right) + {f_x}\left( {a,b} \right)h + {f_y}\left( {a,b} \right)k$
Thus, we obtain the form:
$f\left( {a + h,b + k} \right) \approx a{\left( {1 + b} \right)^{ - 1}} + {\left( {1 + b} \right)^{ - 1}}h - a{\left( {1 + b} \right)^{ - 2}}k$
$f\left( {8 + h,1 + k} \right) \approx 8{\left( {1 + 1} \right)^{ - 1}} + {\left( {1 + 1} \right)^{ - 1}}h - 8{\left( {1 + 1} \right)^{ - 2}}k$
(1) ${\ \ \ \ }$ $f\left( {8 + h,1 + k} \right) \approx 4 + \frac{h}{2} - 2k$
Next, estimate $\frac{{7.98}}{{2.02}}$
We can write $\frac{{7.98}}{{2.02}} = 7.98{\left( {1 + 1.02} \right)^{ - 1}}$.
Let $\left( {x,y} \right) = \left( {7.98,1.02} \right)$. Thus, $\frac{{7.98}}{{2.02}} = 7.98{\left( {1 + 1.02} \right)^{ - 1}} = f\left( {7.98,1.02} \right)$
For $\left( {a,b} \right) = \left( {8,1} \right)$ and $\left( {h,k} \right) = \left( { - 0.02,0.02} \right)$ we have
$x = a + h = 7.98$ ${\ \ }$ and ${\ \ }$ $y = b + k = 1.02$
Using equation (1) we obtain
$\frac{{7.98}}{{2.02}} = f\left( {7.98,1.02} \right) \simeq 4 + \frac{{ - 0.02}}{2} - 2\cdot0.02$
$\frac{{7.98}}{{2.02}} = f\left( {7.98,1.02} \right) \simeq 4 - 0.01 - 0.04$
$\frac{{7.98}}{{2.02}} \simeq 3.95$
Using a calculator: $\frac{{7.98}}{{2.02}} \simeq 3.9505$
