Answer
$$\eqalign{
& y' = \frac{{3x + 1}}{{2\sqrt {x + 2} }} \cr
& y'' = \frac{{3x + 11}}{{4{{\left( {x + 2} \right)}^{3/2}}}} \cr
& y''' = \frac{{6{{\left( {x + 2} \right)}^{1/2}} - 3\left( {3x + 11} \right)}}{{8{{\left( {x + 2} \right)}^{5/2}}}} \cr} $$
Work Step by Step
$$\eqalign{
& y = \left( {x - 3} \right)\sqrt {x + 2} \cr
& {\text{Calculate }}y'{\text{ differentiating both sides with respect to }}x \cr
& y' = \frac{d}{{dx}}\left[ {\left( {x - 3} \right)\sqrt {x + 2} } \right] \cr
& y' = \left( {x - 3} \right)\left( {\frac{1}{{2\sqrt {x + 2} }}} \right) + \sqrt {x + 2} \cr
& y' = \frac{{x - 3}}{{2\sqrt {x + 2} }} + \sqrt {x + 2} \cr
& y' = \frac{{x - 3 + 2\left( {x + 2} \right)}}{{2\sqrt {x + 2} }} \cr
& y' = \frac{{3x + 1}}{{2\sqrt {x + 2} }} \cr
& \cr
& {\text{Calculate }}y''{\text{ differentiating both sides with respect to }}x \cr
& y'' = \frac{d}{{dx}}\left[ {\frac{{3x + 1}}{{2\sqrt {x + 2} }}} \right] \cr
& y'' = \frac{{2\sqrt {x + 2} \left( 3 \right) - \left( {3x + 1} \right)\left( {\frac{1}{{\sqrt {x + 2} }}} \right)}}{{{{\left( {2\sqrt {x + 2} } \right)}^2}}} \cr
& {\text{Simplifying}} \cr
& y'' = \frac{{6\sqrt {x + 2} - \frac{{3x + 1}}{{\sqrt {x + 2} }}}}{{4\left( {x + 2} \right)}} \cr
& y'' = \frac{{6\left( {x + 2} \right) - 3x - 1}}{{4{{\left( {x + 2} \right)}^{3/2}}}} \cr
& y'' = \frac{{6x + 12 - 3x - 1}}{{4{{\left( {x + 2} \right)}^{3/2}}}} \cr
& y'' = \frac{{3x + 11}}{{4{{\left( {x + 2} \right)}^{3/2}}}} \cr
& {\text{Calculate }}y'''{\text{ differentiating both sides with respect to }}x \cr
& y''' = \frac{{4{{\left( {x + 2} \right)}^{3/2}}\left( 3 \right) - \left( {3x + 11} \right)\left( 6 \right){{\left( {x + 2} \right)}^{1/2}}}}{{16{{\left( {x + 2} \right)}^3}}} \cr
& y''' = \frac{{12{{\left( {x + 2} \right)}^{3/2}} - 6\left( {3x + 11} \right){{\left( {x + 2} \right)}^{1/2}}}}{{16{{\left( {x + 2} \right)}^3}}} \cr
& y''' = \frac{{6{{\left( {x + 2} \right)}^{1/2}} - 3\left( {3x + 11} \right)}}{{8{{\left( {x + 2} \right)}^{5/2}}}} \cr} $$
