Derivative of secant and evaluation at a point

Apply the memorized trigonometric derivative rule directly: $p'(x) = \frac{d}{dx}[\sec x] = \sec x \tan x$. This is one of the six standard trig derivatives and requires no further algebraic manipulation.

Now substitute $x = \pi/3$: recall $\sec(\pi/3) = 2$ and $\tan(\pi/3) = \sqrt{3}$. Therefore $p'(\pi/3) = \sec(\pi/3)\tan(\pi/3) = 2\cdot \sqrt{3} = 2\sqrt{3}$.