I've got this function: $$f(x)=\left(\frac{\sec x+\tan x}{\sec x-\tan x}\right)^{1/2}$$ and I tried using pythagorean identity to simplify it, thus I've found that:
$$f(x)=\sec x+\tan x$$
and derivative is way more simple here. Is this OK? or should I consider function domain and stuff?
No, because\begin{align}\frac{\sec x+\tan x}{\sec x-\tan x}&=\frac{(\sec x+\tan x)^2}{(\sec x-\tan x)(\sec x+\tan x)}\\&=(\sec x+\tan x)^2\end{align}and, for each $x\in\Bbb R$ such that $\cos x\ne0$,$$\sec x+\tan x=\frac{1+\sin x}{\cos x}.$$So,$$\sqrt{\frac{\sec x+\tan x}{\sec x-\tan x}}=\begin{cases}\sec x+\tan x&\text{ if }\cos x>0\\-\sec x-\tan x&\text{ otherwise.}\end{cases}$$
You have to consider domain to find $f(x)$ & $f'(x)$, $$f(x)=\sqrt{\frac{\sec x+\tan x}{\sec x-\tan x}}$$ $$=\sqrt{\frac{(\sec x+\tan x)^2}{(\sec x-\tan x)(\sec x+\tan x)}}$$ $$=\left(\frac{(\sec x+\tan x)^2}{sec^2x-\tan^2x}\right)^{1/2}$$ $$=\sqrt{(\sec x+\tan x)^2}$$ $$=|\sec x+\tan x|$$
$$\therefore f(x)=\begin{cases}\sec x+\tan x\quad \forall \ \ x\in\left[2n\pi-\frac{\pi}{2}, 2n\pi+\frac{\pi}{2}\right]\\ -\sec x-\tan x \ \ \quad x\in\left(2n\pi+\frac{\pi}{2}, 2n\pi+\frac{3\pi}{2}\right)\end{cases}$$
Where $n$ is any integer i.e. $n=0, \pm1, \pm2, \pm3, \ldots$
Note that the range of the function stays positive. You should have, instead,
$$f(x) = |\sec x+\tan x|$$
and the corresponding derivative is
$$f’(x) = \sec x |\sec x+\tan x|$$
