Answer
$f\left( {3.01,3.99} \right) = \sqrt {{{3.01}^2} + {{3.99}^2}} \simeq 4.998$
Using a calculator:
$\sqrt {{{3.01}^2} + {{3.99}^2}} \simeq 4.99802$.
Work Step by Step
Let $f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $. So, the partial derivatives are
${f_x} = \frac{x}{{\sqrt {{x^2} + {y^2}} }}$, ${\ \ \ }$ ${f_y} = \frac{y}{{\sqrt {{x^2} + {y^2}} }}$
We can consider $\sqrt {{{3.01}^2} + {{3.99}^2}} $ as a value of $f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $. Thus,
$f\left( {3.01,3.99} \right) = \sqrt {{{3.01}^2} + {{3.99}^2}} $
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 $\left( {a,b} \right) = \left( {3,4} \right)$ and $\left( {h,k} \right) = \left( {0.01, - 0.01} \right)$, we get
$f\left( {3.01,3.99} \right) \approx f\left( {3,4} \right) + {f_x}\left( {3,4} \right)\cdot0.01 + {f_y}\left( {3,4} \right)\cdot\left( { - 0.01} \right)$
$f\left( {3.01,3.99} \right) \simeq 5 + \frac{3}{5}\cdot0.01 - \frac{4}{5}\cdot0.01$
$f\left( {3.01,3.99} \right) = \sqrt {{{3.01}^2} + {{3.99}^2}} \simeq 4.998$
Using a calculator, we get $\sqrt {{{3.01}^2} + {{3.99}^2}} \simeq 4.99802$.
