The full question is: Let $f(x) = x \arcsin\left(\frac{2x}{1+x^2}\right),$ $x \in\mathbb R.$ Find the derived function $f'(x)$ by considering it as a composition of two functions, and provide the domain of the existence of $f'(x)$.
I tried doing it as $g(x)=\tan x$ and figuring out something for $h(x)$ but the boundary conditions for the conversion of the inverse trigonometric functions keep confusing me.
It's simpler than looks. Firstly, by the product rule, if $f(x)=xg(x)$, then: $$f'(x)=xg'(x)+g(x)$$ Now, we have $g(x)=\arcsin(\frac{2x}{1+x^2})$. We set $\alpha(x)=\arcsin(x), \beta(x)=\frac{2x}{1+x^2}$, then $g(x)=\alpha(\beta(x))$, and by the chain rule: $$\frac{d[\alpha(\beta(x))]}{dx}=\frac{d[\beta(x)]}{dx}\cdot\frac{d[\alpha(\beta(x))]}{d[\beta(x)]}\implies(\alpha(\beta))'=\beta'\cdot\alpha'(\beta)$$
We can calculate $\alpha', \beta'$ and force through the first equation to find $f'$: $$f'=x\beta'\alpha'(\beta)+\alpha(\beta)$$