\[\begin{align*} \boxed{\dfrac{d}{dx} \sec(2x) = 2 \cdot \sec \left(2x\right)\tan \left(2x\right)} \end{align*}\]

Solving for the Derivative

Applying the chain rule:

\[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)) \end{align*}\]

We may set our $g(x) = 2x$ and our $f(x) = \sec x$. Doing this, we get that

\[\begin{align*} \dfrac{d}{dx} \sec(2x) &= \sec \left(2x\right)\tan \left(2x\right)\frac{d}{dx}\left(2x\right)\\ &= \boxed{2 \cdot \sec \left(2x\right)\tan \left(2x\right)} \end{align*}\]

This result is immediate from the common derivative of $\sec x$ being $\sec x \tan x$. Assuming we do not know this derivative, the easiest way to show the derivative would be using the quotient rule:

\[\begin{align*} \dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x)f'(x) – f(x)g'(x)}{g^2(x)} \end{align*}\]

Applying this to the derivative of $\sec x$, notice that $\sec x = \dfrac{1}{\cos x}$, so we may say that:

\[\begin{align*} \dfrac{d}{dx} \sec x = \dfrac{d}{dx} \dfrac{1}{\cos x} &= \dfrac{\cos x \cdot 0 – 1 \cdot -\sin x}{\cos^2 x}\\ &= \dfrac{\sin x}{\cos^2 x}\\ &= \dfrac{1}{\cos x} \cdot \dfrac{\sin x}{\cos x}\\ &= \sec x \tan x \end{align*}\]

Note on $\sec x$

It is advised you remember common trigonometric derivatives such as $\sec x$ for an easier time when deriving other problems!