Answer
$5\pi$
Work Step by Step
Step 1. With $u=g(x)$, $(f(g(x))'=f'(u)g'(x)=(1+(-2)cos^{-3}u(-sin(u)))(\pi)=\pi(1+\frac{2sin(u)}{cos^3u})$
Step 2. At $x=1/4$, we have $u=\pi/4$, thus $(f(g(x))'=\pi(1+\frac{2sin(\pi/4)}{cos^3\pi/4})=\pi(1+\sqrt 2\cdot 2\sqrt 2)=5\pi$
