Answer
$f\left( {0.98,2.01} \right) = \frac{{{{0.98}^2}}}{{{{2.01}^3} + 1}} \simeq 0.105185$
Using a calculator:
$\frac{{{{0.98}^2}}}{{{{2.01}^3} + 1}} \simeq 0.1053$.
Work Step by Step
Let $f\left( {x,y} \right) = \frac{{{x^2}}}{{{y^3} + 1}}$. So, the partial derivatives are
${f_x} = \frac{{2x}}{{{y^3} + 1}}$, ${\ \ \ }$ ${f_y} = - \frac{{3{x^2}{y^2}}}{{{{\left( {{y^3} + 1} \right)}^2}}}$
We can consider $\frac{{{{0.98}^2}}}{{{{2.01}^3} + 1}}$ as a value of $f\left( {x,y} \right) = \frac{{{x^2}}}{{{y^3} + 1}}$. Thus,
$f\left( {0.98,2.01} \right) = \frac{{{{0.98}^2}}}{{{{2.01}^3} + 1}}$
Using the linear approximation, Eq. (3) we have
$f\left( {a + h,b + k} \right) \approx f\left( {a,b} \right) + {f_x}\left( {a,b} \right)h + {f_y}\left( {a,b} \right)k$
For $\left( {a,b} \right) = \left( {1,2} \right)$ and $\left( {h,k} \right) = \left( { - 0.02,0.01} \right)$, we get
$f\left( {0.98,2.01} \right) \approx f\left( {1,2} \right) + {f_x}\left( {1,2} \right)\cdot\left( { - 0.02} \right) + {f_y}\left( {1,2} \right)\cdot0.01$
$f\left( {0.98,2.01} \right) \simeq \frac{1}{9} + \frac{2}{9}\cdot\left( { - 0.02} \right) - \frac{4}{{27}}\cdot0.01$
$f\left( {0.98,2.01} \right) = \frac{{{{0.98}^2}}}{{{{2.01}^3} + 1}} \simeq 0.105185$
Using a calculator, we get $\frac{{{{0.98}^2}}}{{{{2.01}^3} + 1}} \simeq 0.1053$.
