$f(x) = \sin {\cos ^{ - 1}}\left( {{{1 - {{({2^x})}^2}} \over {1 + {{({2^x})}^2}}}} \right)$ $= \sin (2{\tan ^{ - 1}}{2^x})$ $f'(x) = \cos (2{\tan ^{ - 1}}{2^x}).2.{1 \over {1 + {{({2^x})}^2}}} \times {2^x}.{\log _e}2$ $\therefore$ $f'(1) = \cos (2{\tan ^{ - 1}}2).{2 \over {1 + 4}} \times 2 \times {\log _e}2$ $\Rightarrow f'(1) = \cos {\cos ^{ - 1}}\left( {{{1 - {2^2}} \over {1 + {2^2}}}} \right).{4 \over 5}{\log _e}2$ $= - {{12} \over {25}}{\log _e}2$ $\Rightarrow a = 25,b = 12$ $|{a^2} - {b^2}| = |625 - 144| = 481$