Answer
$\Delta f = f\left( {2.03,0.9} \right) - f\left( {2,1} \right) \simeq 3.56$
Work Step by Step
We have $f\left( {x,y} \right) = {x^3}{y^{ - 4}}$. The partial derivatives are
${f_x} = 3{x^2}{y^{ - 4}}$, ${\ \ \ }$ ${f_y} = - 4{x^3}{y^{ - 5}}$
For $\left( {a,b} \right) = \left( {2,1} \right)$ and $\left( {\Delta x,\Delta y} \right) = \left( {0.03, - 0.1} \right)$, we get
$\Delta f = f\left( {a + \Delta x,b + \Delta y} \right) - f\left( {a,b} \right) = f\left( {2.03,0.9} \right) - f\left( {2,1} \right)$
Using Eq. (5) we obtain the desired estimate of change
$\Delta f = f\left( {2.03,0.9} \right) - f\left( {2,1} \right) \approx {f_x}\left( {2,1} \right)\Delta x + {f_y}\left( {2,1} \right)\Delta y$
$\Delta f = f\left( {2.03,0.9} \right) - f\left( {2,1} \right) \approx 12\cdot0.03 - 32\cdot\left( { - 0.1} \right)$
$\Delta f \simeq 3.56$
