Answer
$$\frac{{dy}}{{dx}} = \frac{{\sin x\cos x{{\tan }^3}x}}{{\sqrt x }}\left( {\cot x - \tan x + 3\sec x\csc x - \frac{1}{{2x}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{{\sin x\cos x{{\tan }^3}x}}{{\sqrt x }} \cr
& {\text{using radical properties}} \cr
& y = \frac{{\sin x\cos x{{\tan }^3}x}}{{{x^{1/2}}}} \cr
& {\text{taking the natural logarithm on both sides of the equation}} \cr
& \ln y = \ln \left[ {\frac{{\sin x\cos x{{\tan }^3}x}}{{{x^{1/2}}}}} \right] \cr
& {\text{quotient rule of logarithms}} \cr
& \ln y = \ln \left( {\sin x\cos x{{\tan }^3}x} \right) - \ln {x^{1/2}} \cr
& {\text{product rule of logarithms}} \cr
& \ln y = \ln \left( {\sin x} \right) + \ln \left( {\cos x} \right) + \ln \left( {{{\tan }^3}x} \right) - \ln {x^{1/2}} \cr
& {\text{power rule of logarithms}} \cr
& \ln y = \ln \left( {\sin x} \right) + \ln \left( {\cos x} \right) + 3\ln \left( {\tan x} \right) - \frac{1}{2}\ln x \cr
& {\text{Differentiate both sides}} \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {\ln \left( {\sin x} \right)} \right] + \frac{d}{{dx}}\left[ {\ln \left( {\cos x} \right)} \right] + \frac{d}{{dx}}\left[ {3\ln \left( {\tan x} \right)} \right] - \frac{d}{{dx}}\left( {\frac{1}{2}\ln x} \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{\cos x}}{{\sin x}} + \frac{{ - \sin x}}{{\cos x}} + 3\left( {\frac{{{{\sec }^2}x}}{{\tan x}}} \right) - \frac{1}{2}\left( {\frac{1}{x}} \right) \cr
& {\text{Simplify}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \cot x - \tan x + 3\sec x\csc x - \frac{1}{{2x}} \cr
& {\text{Solve for dy/dx}} \cr
& \frac{{dy}}{{dx}} = y\left[ {\cot x - \tan x + 3\sec x\csc x - \frac{1}{{2x}}} \right] \cr
& {\text{substituting }}y = \frac{{\sin x\cos x{{\tan }^3}x}}{{\sqrt x }} \cr
& \frac{{dy}}{{dx}} = \frac{{\sin x\cos x{{\tan }^3}x}}{{\sqrt x }}\left( {\cot x - \tan x + 3\sec x\csc x - \frac{1}{{2x}}} \right) \cr} $$
