If $$f(x) = \sin \left( {{{\cos }^{ - 1}}\left( {{{1 - {2^{2x}}} \over {1 + {2^{2x}}}}} \right)} \right)$$ and its first derivative with respect to x is $- {b \over a}{\log _e}2$ when x = 1, where a and b are integers, then the minimum value of | a2 $-$ b2 | is ____________ .
Answer (integer)
481
Solution
$$f(x) = \sin {\cos ^{ - 1}}\left( {{{1 - {{({2^x})}^2}} \over {1 + {{({2^x})}^2}}}} \right)$$<br><br>$= \sin (2{\tan ^{ - 1}}{2^x})$<br><br>$$f'(x) = \cos (2{\tan ^{ - 1}}{2^x}).2.{1 \over {1 + {{({2^x})}^2}}} \times {2^x}.{\log _e}2$$<br><br>$\therefore$ $f'(1) = \cos (2{\tan ^{ - 1}}2).{2 \over {1 + 4}} \times 2 \times {\log _e}2$<br><br>$$ \Rightarrow f'(1) = \cos {\cos ^{ - 1}}\left( {{{1 - {2^2}} \over {1 + {2^2}}}} \right).{4 \over 5}{\log _e}2$$<br><br>$= - {{12} \over {25}}{\log _e}2$<br><br>$\Rightarrow a = 25,b = 12$<br><br>$|{a^2} - {b^2}| = |625 - 144| = 481$
About this question
Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions
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