$y=\tan ^{-1} \frac{2 x}{1-x^{2}}$ হলে, $\frac{d y}{d x}$ সমান-

$\frac{2}{1-x^{2}}$

$\frac{2}{\sqrt{1+x^{2}}}$

$\frac{2}{\sqrt{1-x^{2}}}$

$\frac{2}{1+x^{2}}$