Answer
$$\frac{{dy}}{{dx}} = \frac{{\sin 3x\cos x - 3\sin x\cos 3x}}{{2\sin 3x\sqrt {\sin x\sin 3x} }}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {\frac{{\sin x}}{{\sin 3x}}} \cr
& {\text{using the property of radicals }}\sqrt u = {u^{1/2}} \cr
& y = {\left( {\frac{{\sin x}}{{\sin 3x}}} \right)^{1/2}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}{\left( {\frac{{\sin x}}{{\sin 3x}}} \right)^{1/2}} \cr
& {\text{use the general power rule for derivatives }}{D_x}\left[ {{u^n}} \right] = n{u^{n - 1}}{D_x}\left( u \right).{\text{ consider }}u = \frac{{\sin x}}{{2\sin 3x}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\frac{{\sin x}}{{\sin 3x}}} \right)^{ - 1/2}}{D_x}\left( {\frac{{\sin x}}{{\sin 3x}}} \right) \cr
& {\text{By using the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\frac{{\sin 3x}}{{\sin x}}} \right)^{1/2}}\frac{{\sin 3x \cdot {D_x}\left( {\sin x} \right) - \sin x \cdot {D_x}\left( {\sin 3x} \right)}}{{{{\left( {\sin 3x} \right)}^2}}} \cr
& {\text{solve the derivatives}}{\text{, use }}{D_x}\left( {\sin u} \right) = \cos u \cdot {D_x}\left( u \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\frac{{\sin 3x}}{{\sin x}}} \right)^{1/2}}\frac{{\sin 3x\left( {\cos x} \right) - \sin x\left( {3\cos 3x} \right)}}{{{{\left( {\sin 3x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\frac{{\sin 3x}}{{\sin x}}} \right)^{1/2}}\frac{{\sin 3x\cos x - 3\sin x\cos 3x}}{{{{\left( {\sin 3x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{{{{\left( {\sin x} \right)}^{1/2}}}}} \right)\frac{{\sin 3x\cos x - 3\sin x\cos 3x}}{{{{\left( {\sin 3x} \right)}^{3/2}}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\sin 3x\cos x - 3\sin x\cos 3x}}{{2\sin 3x\sqrt {\sin x} \sqrt {\sin 3x} }} \cr
& \frac{{dy}}{{dx}} = \frac{{\sin 3x\cos x - 3\sin x\cos 3x}}{{2\sin 3x\sqrt {\sin x\sin 3x} }} \cr} $$
