If \(\mathrm{f}(x)=\sin ^{-1}\left(\frac{2 \cdot 3^x}{1+9^x}\right)\), then \(\mathrm{f}^{\prime}\left(\frac{1}{2}\right)\) equals

\(\begin{aligned} & \mathrm{f}(x)=\sin ^{-1}\left(\frac{2.3^x}{1+9^x}\right)=\sin ^{-1}\left(\frac{2.3^x}{1+\left(3^x\right)^2}\right) \\ & \text { Put } 3^x=\tan \theta \Rightarrow \theta=\tan ^{-1}\left(3^x\right) \\ & \therefore \quad \mathrm{f}(x)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \\ &=\sin ^{-1}(\sin 2 \theta) \\ &=2 \theta\end{aligned}\)
\(\begin{aligned} \therefore & \mathrm{f}(x)=2 \tan ^{-1}\left(3^x\right) \\ \therefore & \mathrm{f}^{\prime}(x)=2 \cdot \frac{1}{1+\left(3^x\right)^2} \cdot 3^x \log 3 \\ \therefore \quad \mathrm{f}^{\prime}\left(\frac{1}{2}\right) & =2 \cdot \frac{1}{1+\left(3^{\frac{1}{2}}\right)^2} \cdot 3^{\frac{1}{2}} \log 3 \\ & =\frac{1}{2} \sqrt{3} \log 3=\sqrt{3} \log \sqrt{3}\end{aligned}\)