Answer
The derivative is $\frac{(sin(x)+2xcos(x))(\sqrt x)}{2x}$.
Work Step by Step
Product Rule $g'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=\sqrt x ;u’(x)=\frac{1}{2\sqrt x} $
$v(x)=\sin(x) ;v’(x)=\cos(x)$
$g'(x)=(\frac{1}{2\sqrt x})(\sin(x))+(\sqrt x)(cos(x))=$
$(\frac{1}{2\sqrt x})(\sin(x))+(\frac{2x}{2\sqrt x})(cos(x))=$
$(\frac{1}{2\sqrt x})(sin(x)+2xcos(x))=$
$\frac{sin(x)+2xcos(x)}{2\sqrt x}=$
$\frac{(sin(x)+2xcos(x))(\sqrt x)}{2x}$
