Answer
$f\left( {8.01,1.99,2.01} \right) = \frac{{8.01}}{{\sqrt {\left( {1.99} \right)\left( {2.01} \right)} }} \simeq 4.005$
Using a calculator:
$\frac{{8.01}}{{\sqrt {\left( {1.99} \right)\left( {2.01} \right)} }} \simeq 4.00505$.
Work Step by Step
Let $f\left( {x,y,z} \right) = \frac{x}{{\sqrt {yz} }}$. So, the partial derivatives are
${f_x} = \frac{1}{{\sqrt {yz} }}$, ${\ \ \ }$ ${f_y} = - \frac{{xz}}{{2{{\left( {yz} \right)}^{3/2}}}}$, ${\ \ \ }$ ${f_z} = - \frac{{xy}}{{2{{\left( {yz} \right)}^{3/2}}}}$
We can consider $\frac{{8.01}}{{\sqrt {\left( {1.99} \right)\left( {2.01} \right)} }}$ as a value of $f\left( {x,y} \right) = \frac{x}{{\sqrt {yz} }}$. Thus,
$f\left( {8.01,1.99,2.01} \right) = \frac{{8.01}}{{\sqrt {\left( {1.99} \right)\left( {2.01} \right)} }}$
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 $\left( {a,b,c} \right) = \left( {8,2,2} \right)$ and $\left( {h,k,m} \right) = \left( {0.01, - 0.01,0.01} \right)$, we get
$f\left( {8.01,1.99,2.01} \right) \approx f\left( {8,2,2} \right) + {f_x}\left( {8,2,2} \right)\cdot0.01 + {f_y}\left( {8,2,2} \right)\cdot\left( { - 0.01} \right) + {f_z}\left( {8,2,2} \right)\cdot0.01$
$f\left( {8.01,1.99,2.01} \right) \simeq 4 + \frac{1}{2}\cdot0.01 - 1\cdot\left( { - 0.01} \right) - 1\cdot0.01$
$f\left( {8.01,1.99,2.01} \right) = \frac{{8.01}}{{\sqrt {\left( {1.99} \right)\left( {2.01} \right)} }} \simeq 4.005$
Using a calculator, we get $\frac{{8.01}}{{\sqrt {\left( {1.99} \right)\left( {2.01} \right)} }} \simeq 4.00505$.
