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