The derivative of f(tan x) with respect to g(sec x) at $x=\frac{\pi}{4}$, where $f'(1)=2$ and $g'(\sqrt{2})=4$, is

$\frac{1}{\sqrt{2}}$

Let $u=f(\tan x)$ and $v=g(\sec x)$. Then,

$\frac{d u}{d v} =\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{f'(\tan x) \sec ^2 x}{g'(\sec x) \sec x \tan x}$

$\Rightarrow \frac{d u}{d v} =\frac{f'(\tan x)}{g'(\sec x)} ~cosec x$

$\Rightarrow \left(\frac{d u}{d v}\right)_{x=\frac{\pi}{4}}=\sqrt{2} \frac{f'(1)}{g'(\sqrt{2})}=\frac{2 \sqrt{2}}{4}=\frac{1}{\sqrt{2}}$