If f(x) = sin^{-1}left(frac{2 cdot 3^x}{1+9^x | vedclass

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(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+(3^x)^2}ight)$.Let $3^x = 	an 	heta$,then $	heta = 	an^{-1}(3^x)$.Substituting this into t

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$\sqrt{3} \log(\sqrt{3})$

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(A) Given $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+(3^x)^2}ight)$.Let $3^x = 	an 	heta$,then $	heta = 	an^{-1}(3^x)$.Substituting this into the function,we get $f(x) = \sin^{-1}\left(\frac{2 	an 	heta}{1+	an^2 	heta}ight) = \sin^{-1}(\sin 2	heta) = 2	heta = 2 	an^{-1}(3^x)$.Differentiating with respect to $x$,we get $f^{\prime}(x) = 2 \cdot \frac{1}{1+(3^x)^2} \cdot \frac{d}{dx}(3^x) = \frac{2 \cdot 3^x \log 3}{1+9^x}$.Evaluating at $x = \frac{1}{2}$,we have $f^{\prime}\left(\frac{1}{2}ight) = \frac{2 \cdot 3^{1/2} \log 3}{1+9^{1/2}} = \frac{2 \sqrt{3} \log 3}{1+3} = \frac{2 \sqrt{3} \log 3}{4} = \frac{\sqrt{3} \log 3}{2} = \sqrt{3} \log(3^{1/2}) = \sqrt{3} \log(\sqrt{3})$.

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

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