## Calculus 10th Edition

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