Answer
$f(x)=g(x)-\displaystyle \frac{\pi}{2},\quad
x\geq 0$
Work Step by Step
$\displaystyle \frac{d}{dx}[f(x)]=\frac{d}{dx}[\sin^{-1}u],\quad u=\frac{x-1}{x+1}$
$=\displaystyle \frac{1}{\sqrt{1-u^{2}}}\cdot\frac{du}{dx}$
$=\displaystyle \frac{1}{\sqrt{1-(\frac{x-1}{x+1})^{2}}}\cdot\frac{1(x+1)-(x-1)(1)}{(x+1)^{2}}$
$=\displaystyle \frac{2}{(x+1)^{2}\sqrt{\frac{(x+1)^{2}-(x-1)^{2}}{(x+1)^{2}}}}\qquad $... x is nonnegative, so $\sqrt{(x+1)^{2}}=(x+1)$
$=\displaystyle \frac{2}{(x+1)\sqrt{2x+2x}}$
$=\displaystyle \frac{2}{(x+1)2\sqrt{x}}$
$=\displaystyle \frac{1}{(x+1)\sqrt{x}}$
$\displaystyle \frac{d}{dx}[g(x)]=\frac{d}{dx}[2\tan^{-1}u],\quad u=\sqrt{x}$
$=2\displaystyle \cdot\frac{1}{1+u^{2}}\cdot\frac{du}{dx}$
$=\displaystyle \frac{2}{1+(\sqrt{x})^{2}}\cdot\frac{1}{2}x^{-1/2}$
$=\displaystyle \frac{1}{(x+1)\sqrt{x}}$
The functions have equal derivatives, which means that they have graphs vertically shifted in regard to each other.
This means that
$f(x)=g(x)+C$
Set x=0 (to find C)
$\displaystyle \sin^{-1}(\frac{0-1}{0+1})=2\tan^{-1}\sqrt{1}+C$
$\sin^{-1}(-1)=2\tan^{-1}1+C$
... $y=\sin^{-1}x$ is the number in $[-\pi/2, \pi/2]$ for which $\sin y=x.$
... $y=\tan^{-1}x$ is the number in $(-\pi/2, \pi/2)$ for which $\tan y=x.$
$-\displaystyle \frac{\pi}{2}=0+C$
$C=-\displaystyle \frac{\pi}{2}$
So,
$f(x)=g(x)-\displaystyle \frac{\pi}{2},\quad
x\geq 0$
