Derivatives need practice

$\frac{d}{dx}(\sin2x) = 2\cos2x$

$\frac{d}{dx}( \tan2x) = 2 \sec^2 2x$

According to the rule:

if $y= f.g$

then $\frac{dy}{dx}=f`.g+f.g`$

Therefore:

$\frac { d }{ dx } (\sin { 2x } \tan { 2x } ) = 2\cos2x \times \tan2x + \sin2x \times 2 \sec^2 2x$

put $x = \frac{\pi}{3} = 60°$

$\ 2\cos120° \times \tan120° + \sin120° \times 2 \sec^2 120° = \boxed{5\sqrt{3}}$

$tan2x= \frac{sin2x}{cos2x}$

hence $\frac{sin^{2}2x}{cos 2x}$ =

$\frac{1-cos^{2}2x }{cos 2x}= sec 2x -cos 2x$

differentiate it and get sec2x tan2x + sin2x put the value of x and get the answer