Answer
$$\frac{{dy}}{{dx}} = \frac{{\cos 4x\sin x - 4\cos x\sin 4x}}{{2\cos x\sqrt {\cos 4x\cos x} }}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {\frac{{\cos 4x}}{{\cos x}}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}{\left( {\frac{{\cos 4x}}{{\cos x}}} \right)^{1/2}} \cr
& {\text{use the general power rule for derivatives }}{D_x}\left[ {{u^n}} \right] = n{u^{n - 1}}{D_x}\left( u \right).{\text{ }} \cr
& {\text{consider }}u = \frac{{\cos 4x}}{{\cos x}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\frac{{\cos 4x}}{{\cos x}}} \right)^{ - 1/2}}{D_x}\left( {\frac{{\cos 4x}}{{\cos x}}} \right) \cr
& {\text{By using the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\frac{{\cos 4x}}{{\cos x}}} \right)^{ - 1/2}}\frac{{\cos x \cdot {D_x}\left( {\cos 4x} \right) - \cos 4x \cdot {D_x}\left( {\cos x} \right)}}{{{{\left( {\cos x} \right)}^2}}} \cr
& {\text{solve the derivatives}}{\text{, use }}{D_x}\left( {\cos u} \right) = - \sin u \cdot {D_x}\left( u \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\frac{{{{\left( {\cos 4x} \right)}^{ - 1/2}}}}{{{{\left( {\cos x} \right)}^{ - 1/2}}}}\frac{{\cos x\left( { - 4\sin 4x} \right) - \cos 4x\left( { - \sin x} \right)}}{{{{\left( {\cos x} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\frac{{{{\left( {\cos 4x} \right)}^{ - 1/2}}}}{{{{\left( {\cos x} \right)}^{ - 1/2}}}}\frac{{ - 4\cos x\sin 4x + \cos 4x\sin x}}{{{{\left( {\cos x} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{\cos 4x\sin x - 4\cos x\sin 4x}}{{2{{\left( {\cos 4x} \right)}^{1/2}}{{\left( {\cos x} \right)}^{3/2}}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\cos 4x\sin x - 4\cos x\sin 4x}}{{2\cos x\sqrt {\cos 4x\cos x} }} \cr} $$
