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