Answer
$f\left( {1.9,2.02,4.05} \right) = \sqrt {\left( {1.9} \right)\left( {2.02} \right)\left( {4.05} \right)} \simeq 3.945$
Using a calculator:
$\sqrt {\left( {1.9} \right)\left( {2.02} \right)\left( {4.05} \right)} \simeq 3.94258$.
Work Step by Step
Let $f\left( {x,y,z} \right) = \sqrt {xyz} $. So, the partial derivatives are
${f_x} = \frac{{yz}}{{2\sqrt {xyz} }}$, ${\ \ \ }$ ${f_y} = \frac{{xz}}{{2\sqrt {xyz} }}$, ${\ \ \ }$ ${f_z} = \frac{{xy}}{{2\sqrt {xyz} }}$
We can consider $\sqrt {\left( {1.9} \right)\left( {2.02} \right)\left( {4.05} \right)} $ as a value of $f\left( {x,y} \right) = \sqrt {xyz} $. Thus,
$f\left( {1.9,2.02,4.05} \right) = \sqrt {\left( {1.9} \right)\left( {2.02} \right)\left( {4.05} \right)} $
Using the linear approximation in three variables, we have
$f\left( {a + h,b + k,c + m} \right) \approx f\left( {a,b,c} \right) + {f_x}\left( {a,b,c} \right)h + {f_y}\left( {a,b,c} \right)k + {f_z}\left( {a,b,c} \right)m$
For $\left( {a,b,c} \right) = \left( {2,2,4} \right)$ and $\left( {h,k,m} \right) = \left( { - 0.1,0.02,0.05} \right)$, we get
$f\left( {1.9,2.02,4.05} \right) \approx f\left( {2,2,4} \right) + {f_x}\left( {2,2,4} \right)\cdot\left( { - 0.1} \right) + {f_y}\left( {2,2,4} \right)\cdot0.02 + {f_z}\left( {2,2,4} \right)\cdot0.05$
$f\left( {1.9,2.02,4.05} \right) \simeq 4 + 1\cdot\left( { - 0.1} \right) + 1\cdot0.02 + \frac{1}{2}\cdot0.05$
$f\left( {1.9,2.02,4.05} \right) = \sqrt {\left( {1.9} \right)\left( {2.02} \right)\left( {4.05} \right)} \simeq 3.945$
Using a calculator, we get $\sqrt {\left( {1.9} \right)\left( {2.02} \right)\left( {4.05} \right)} \simeq 3.94258$.
