Answer
For $x\gt0,\quad f(x)=g(x)$
Work Step by Step
$\displaystyle \frac{d}{dx}[f(x)]=\frac{d}{dx}[\sin^{-1}u],\quad u=(x^{2}+1)^{-1/2}$
$=\displaystyle \frac{1}{\sqrt{1-u^{2}}}\cdot\frac{du}{dx}$
$=\displaystyle \frac{1}{\sqrt{1-[(x^{2}+1)^{-1/2}]^{2}}}\cdot(-\frac{1}{2}(x^{2}+1)^{-3/2}\cdot 2x)$
$= - \displaystyle \frac{x}{\sqrt{(x^{2}+1)^{3}(1-\frac{1}{x^{2}+1})}}$
$= - \displaystyle \frac{x}{(x^{2}+1)\sqrt{x^{2}+1-1}}$
$= - \displaystyle \frac{x}{(x^{2}+1)\sqrt{x^{2}}}$
$= - \displaystyle \frac{x}{(x^{2}+1)|x|}$
For $x\displaystyle \gt0,\quad \frac{d}{dx}[f(x)]=-\frac{1}{x^{2}+1}$
$\displaystyle \frac{d}{dx}[f(x)]=\frac{d}{dx}[\tan^{-1}u],\quad u=x^{-1}$
$=\displaystyle \frac{1}{1+u^{2}}\cdot\frac{du}{dx}$
$=\displaystyle \frac{1}{1+x^{-2}}\cdot(-x^{-2})$
$=-\displaystyle \frac{1}{x^{2}(1+x^{-2})}$
$=-\displaystyle \frac{1}{x^{2}+1}$
So, for $x\displaystyle \gt0,\quad \frac{d}{dx}[f(x)]= \frac{d}{dx}[g(x)]$
This means that $f(x)=g(x)+C$
To find C, set x=1
$\sin^{-1}$($\displaystyle \frac{1}{\sqrt{1+1^{2}}})=\tan(\frac{1}{1})+C$
$\displaystyle \sin^{-1}(\frac{1}{\sqrt{2}})=\tan^{-1}1+C$
$\displaystyle \frac{\pi}{4}=\frac{\pi}{4}+C$
$C=0$
So, for $x\gt0,\quad f(x)=g(x)$
