Deriving trigonometric function

The differentiation you need is direct from the following formulas:

1. $d(\tan u) = \sec^2 u ~ du$
2. $d(\sec u) = \sec u \tan u ~ du$

From your equation, $u = 2x$ from which $du = 2$, hence,
$\dfrac{d}{dx} \left[ \tan 2x + \sec 2x \right] = 2 \sec^2 2x + 2 \sec 2x \tan 2x$