Since f(x) = sin$$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right)$$

Suppose 3^x = tan t

$$ \Rightarrow $$   f(x) = sin^$$-$$1 $$\left( {{{2\tan t} \over {1 + {{\tan }^2}t}}} \right)$$ = sin^$$-$$1 (sin2t) = 2t

                                         = 2tan^$$-$$1 (3x)

So, f'(x) = $${2 \over {1 + {{\left( {{3^x}} \right)}^2}}} \times {3^x}.{\log _e}3$$

$$ \therefore $$   f '$$\left( { - {1 \over 2}} \right)$$ = $${2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}}$$ . log_e 3

=  $${1 \over 2} \times \sqrt 3 \times {\log _e}3$$   =  $$\sqrt 3 \times {\log _e}\sqrt 3 $$