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

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$\begin{aligned} & \mathrm{f}(x)=\sin ^{-1}\left(\frac{2.3^x}{1+9^x}ight)=\sin ^{-1}\left(\frac{2.3^x}{1+\left(3^xight)^2}ight) \ & 	ext { Put } 3^x=	an 	heta \Rightarrow 	heta=	an ^{-1}\left(3^xight) \ & 	herefore \quad \mathrm{f}(x)=\sin ^{-1}\left(\frac{2 	an 	heta}{1+	an ^2 	heta}ight) \ &=\sin ^{-1}(\sin 2 	heta) \ &=2 	heta\end{aligned}$ $\begin{aligned} 	herefore & \mathrm{f}(x)=2 	an ^{-1}\left(3^xight) \ 	herefore & \mathrm{f}^{\prime}(x)=2 \cdot \frac{1}{1+\left(3^xight)^2} \cdot 3^x \log 3 \ 	herefore \quad \mathrm{f}^{\prime}\left(\frac{1}{2}ight) & =2 \cdot \frac{1}{1+\left(3^{\frac{1}{2}}ight)^2} \cdot 3^{\frac{1}{2}} \log 3 \ & =\frac{1}{2} \sqrt{3} \log 3=\sqrt{3} \log \sqrt{3}\end{aligned}$

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

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