Answer
$\sqrt {9.1/3.9} \simeq 1.52708$
Work Step by Step
We are given $f\left( {x,y} \right) = \sqrt {x/y} $. The partial derivatives are
${f_x} = \frac{1}{{2\sqrt {xy} }}$, ${\ \ \ }$ ${f_y} = - \frac{{\sqrt x }}{{2{y^{3/2}}}}$
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 $\left( {a,b} \right) = \left( {9,4} \right)$ and $\left( {h,k} \right) = \left( {0.1, - 0.1} \right)$ we have
$x = a + h = 9.1$ ${\ \ }$ and ${\ \ }$ $y = b + k = 3.9$
Thus,
$f\left( {9.1,3.9} \right) = \sqrt {9.1/3.9} \approx f\left( {9,4} \right) + {f_x}\left( {9,4} \right)\cdot0.1 + {f_y}\left( {9,4} \right)\cdot\left( { - 0.1} \right)$
$\sqrt {9.1/3.9} \simeq \frac{3}{2} + \frac{1}{{12}}\cdot0.1 + \left( { - \frac{3}{{16}}} \right)\cdot\left( { - 0.1} \right)$
$\sqrt {9.1/3.9} \simeq 1.52708$
