Answer
$$\Delta y \approx 0.37$$
Work Step by Step
$$\eqalign{
& y = x\sqrt {8x + 1} ;{\text{ from }}x = 3{\text{ to }}x = 3.05 \cr
& {\text{Calculate }}f'\left( x \right){\text{ and }}dx \cr
& f\left( x \right) = x\sqrt {8x + 1} \cr
& f'\left( x \right) = x\left( {\frac{8}{{2\sqrt {8x + 1} }}} \right) + \sqrt {8x + 1} \cr
& f'\left( x \right) = \frac{{4x}}{{\sqrt {8x + 1} }} + \sqrt {8x + 1} \cr
& dx = \Delta x \cr
& dx = 3.05 - 3 \cr
& dx = 0.05 \cr
& \cr
& \Delta y \approx f'\left( x \right)dx = dy \cr
& \Delta y \approx \left[ {\frac{{4x}}{{\sqrt {8x + 1} }} + \sqrt {8x + 1} } \right]dx \cr
& {\text{Substitute }}dx = - 0.04{\text{ and }}x = 3 \cr
& \Delta y \approx \left[ {\frac{{4\left( 3 \right)}}{{\sqrt {8\left( 3 \right) + 1} }} + \sqrt {8\left( 3 \right) + 1} } \right]\left( {0.05} \right) \cr
& \Delta y \approx \left( {\frac{{12}}{5} + 5} \right)\left( {0.05} \right) \cr
& \Delta y \approx 0.37 \cr} $$
