#### Answer

For $x=1/4$, $$(fog)'=5\pi$$

#### Work Step by Step

$f(u)=u+\frac{1}{\cos^2u}$, $u=g(x)=\pi x$, $x=1/4$
Applying the Chain Rule: $$(fog)'=f'(g(x))g'(x)=f'(u)u'$$
We have
$$f'(u)=\Big(u+\frac{1}{\cos^2u}\Big)'=1+\Big(\frac{-1(\cos^2u)'}{\cos^4u}\Big)=1-\frac{2\cos u(\cos u)'}{\cos^4u}$$ $$f'(u)=1+\frac{2\cos u\sin u}{\cos^4u}=1+\frac{2\sin u}{\cos^3u}$$
$$u'=(\pi x)'=\pi$$
Therefore,
$$(fog)'=\Big(1+\frac{2\sin u}{\cos^3 u}\Big)\pi=\Big(1+\frac{2\sin \pi x}{\cos^3 \pi x}\Big)\pi$$
For $x=1/4$, $$(fog)'=\Big(1+\frac{2\sin\frac{\pi}{4}}{\cos^3\frac{\pi}{4}}\Big)\pi=\Big(1+\frac{2\times\frac{\sqrt2}{2}}{\Big(\frac{\sqrt2}{2}\Big)^3}\Big)\pi=\Big(1+\frac{\sqrt2}{\frac{2\sqrt2}{8}}\Big)\pi$$
$$=\Big(1+\frac{8\sqrt2}{2\sqrt2}\Big)\pi=(1+4)\pi=5\pi$$