Answer
For $\left( {a,b} \right) = \left( {4,1} \right)$:
$f\left( {4.01,0.98} \right) \simeq 8.2$
Work Step by Step
We are given $f\left( {x,y} \right) = \frac{{{x^2}}}{{{y^2} + 1}}$. The partial derivatives are
${f_x} = \frac{{2x}}{{{y^2} + 1}}$, ${\ \ \ }$ ${f_y} = - \frac{{2{x^2}y}}{{{{\left( {{y^2} + 1} \right)}^2}}}$
According to Eq. (3), the linear approximation to $f\left( {x,y} \right)$ is given by
$f\left( {a + h,b + k} \right) \approx f\left( {a,b} \right) + {f_x}\left( {a,b} \right)h + {f_y}\left( {a,b} \right)k$
For $\left( {a,b} \right) = \left( {4,1} \right)$ and $\left( {h,k} \right) = \left( {0.01, - 0.02} \right)$ we have
$x = a + h = 4.01$ ${\ \ }$ and ${\ \ }$ $y = b + k = 0.98$
Thus,
$f\left( {4.01,0.98} \right) \approx f\left( {4,1} \right) + {f_x}\left( {4,1} \right)\cdot0.01 + {f_y}\left( {4,1} \right)\cdot\left( { - 0.02} \right)$
$f\left( {4.01,0.98} \right) \simeq 8 + 4\cdot0.01 + 8\cdot0.02 = 8.2$
