Official solution
Since f(x) = sin\[\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right)\]
Suppose 3x = 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 '\[\left( { - {1 \over 2}} \right)\] = \[{2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}}\] . loge 3
= \[{1 \over 2} \times \sqrt 3 \times {\log _e}3\] = \[\sqrt 3 \times {\log _e}\sqrt 3 \]