Answer
$y'=2\sec^2{(2x)}\cos{(\tan{2x})}.$
Work Step by Step
$u=\tan{2x}$; $\dfrac{du}{dx}=2\sec^2{2x}.$
$\dfrac{dy}{du}=\cos{u}$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=(2\sec^2{2x})(\cos({\tan{2x})})$
$=2\sec^2{(2x)}\cos{(\tan{2x})}.$
