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

We have the function $$f(x)=\sin^{-1}\!\left(\dfrac{2\,3^{x}}{1+9^{x}}\right).$$ To differentiate it, we first note the standard derivative formula

$$\dfrac{d}{dx}\bigl[\sin^{-1}(u)\bigr]=\dfrac{u'}{\sqrt{1-u^{2}}},$$

where $$u=u(x).$$ So we set

$$u(x)=\dfrac{2\,3^{x}}{1+9^{x}}=\dfrac{2\,3^{x}}{1+3^{2x}}.$$

Now we compute $$u'$$ by the quotient rule, which states

$$\dfrac{d}{dx}\!\left(\dfrac{N}{D}\right)=\dfrac{N'D-ND'}{D^{2}},$$

where $$N=2\,3^{x}$$ and $$D=1+3^{2x}.$$

First we differentiate the numerator and the denominator separately.

For the numerator: $$N=2\,3^{x}\quad\Longrightarrow\quad N'=2\,3^{x}\ln 3.$$

For the denominator: $$D=1+3^{2x}\quad\Longrightarrow\quad D'=3^{2x}\cdot 2\ln 3=2\,3^{2x}\ln 3.$$

Substituting into the quotient rule gives

$$u'(x)=\dfrac{\bigl(2\,3^{x}\ln 3\bigr)(1+3^{2x})-\bigl(2\,3^{x}\bigr)\bigl(2\,3^{2x}\ln 3\bigr)}{(1+3^{2x})^{2}}.$$

We now evaluate every quantity at $$x=-\dfrac12.$$ We have $$3^{x}=3^{-1/2}=\dfrac1{\sqrt3},\qquad 3^{2x}=3^{-1}= \dfrac13.$$ Hence

$$N\Bigl(-\tfrac12\Bigr)=2\cdot\dfrac1{\sqrt3}=\dfrac2{\sqrt3},\quad N'\Bigl(-\tfrac12\Bigr)=2\cdot\dfrac1{\sqrt3}\ln 3=\dfrac{2\ln3}{\sqrt3},$$ $$D\Bigl(-\tfrac12\Bigr)=1+\dfrac13=\dfrac43,\quad D'\Bigl(-\tfrac12\Bigr)=2\cdot\dfrac13\ln 3=\dfrac{2\ln3}{3}.$$

So

$$u'\!\Bigl(-\tfrac12\Bigr)= \dfrac{\displaystyle\left(\dfrac{2\ln3}{\sqrt3}\right)\!\left(\dfrac43\right)-\left(\dfrac2{\sqrt3}\right)\!\left(\dfrac{2\ln3}{3}\right)} {\left(\dfrac43\right)^{2}} =\dfrac{\dfrac{8\ln3}{3\sqrt3}-\dfrac{4\ln3}{3\sqrt3}}{\dfrac{16}{9}} =\dfrac{\dfrac{4\ln3}{3\sqrt3}}{\dfrac{16}{9}} =\dfrac{4\ln3}{3\sqrt3}\cdot\dfrac{9}{16} =\dfrac{36\ln3}{48\sqrt3} =\dfrac{3\ln3}{4\sqrt3} =\dfrac{\sqrt3\,\ln3}{4}. $$

Next we need $$\sqrt{1-u^{2}}$$ at $$x=-\dfrac12.$$ First find $$u\bigl(-\tfrac12\bigr):$$

$$u\!\Bigl(-\tfrac12\Bigr)=\dfrac{2\cdot\dfrac1{\sqrt3}}{1+\dfrac13} =\dfrac{\dfrac2{\sqrt3}}{\dfrac43} =\dfrac2{\sqrt3}\cdot\dfrac34 =\dfrac6{4\sqrt3} =\dfrac3{2\sqrt3} =\dfrac{\sqrt3}{2}. $$

Therefore

$$1-u^{2} =1-\left(\dfrac{\sqrt3}{2}\right)^{2} =1-\dfrac34 =\dfrac14,$$ $$\sqrt{1-u^{2}}=\sqrt{\dfrac14}=\dfrac12.$$

Finally, using the derivative formula for $$\sin^{-1}u,$$ we obtain

$$f'\!\Bigl(-\tfrac12\Bigr) =\dfrac{u'\!\left(-\tfrac12\right)}{\sqrt{1-u^{2}}} =\dfrac{\dfrac{\sqrt3\,\ln3}{4}}{\dfrac12} =\dfrac{\sqrt3\,\ln3}{4}\times2 =\dfrac{\sqrt3\,\ln3}{2} =\sqrt3\;\ln\!\sqrt3.$$

The last equality follows because $$\ln\!\sqrt3=\dfrac12\ln 3,$$ so $$\sqrt3\,\ln\!\sqrt3=\sqrt3\cdot\dfrac12\ln3=\dfrac{\sqrt3\,\ln3}{2},$$ exactly the value we obtained.

Thus, $$f'\!\Bigl(-\dfrac12\Bigr)=\sqrt{3}\,\log_{e}\sqrt{3}.$$

Hence, the correct answer is Option A.