$y=\sin ^{-1}\left(\frac{e^{a x}-e^{-a x}}{e^{a x}+e^{-a x}}\right)$, then $\frac{d y}{d x}$ is

$\frac{2 a}{e^{a x}+e^{-a x}}$

$y=\sin ^{-1}\left(\frac{e^{a x}-e^{-a x}}{e^{a x}+e^{-a x}}\right)$,

$\frac{d y}{d x}=\frac{1}{\sqrt{1-\frac{\left(e^{a x}-e^{-a x}\right)^2}{\left(e^{a x}+a^{-a x}\right)^2}}} \times\left\{\frac{a\left(e^{a x}+e^{-a x}\right)}{\left(e^{a x}+e^{-a x}\right)}+\frac{\left(e^{a x}-e^{-a x}\right) \times-1 \times a\left(e^{a x}-e^{-a x}\right)}{\left(e^{a x}+e^{-a x}\right)^2}\right\}$

$=\frac{\left(e^{a x}+e^{-a x}\right)}{\sqrt{e^{2 a x}+y^{-2 a x}+2-e^{2 a x}-e^{-2 a x}+2}} a\left\{\frac{\left(e^{a x}+e^{-a x}\right)^2-\left(e^{a x}-e^{-a x}\right)^2}{\left(e^{a x}+e^{-a x}\right)^2}\right\}$

$=\frac{\frac{a}{2}(2+2)}{\left(e^{a x}+e^{-a x}\right)}=\frac{2 a}{e^{a x}+e^{-a x}}$

Hence (2) is correct answer.