Hàm số (fleft( x ight) = {log _2}left( {{2^x} + sqrt {{4^x} + 1} } ight)) có đạo hàm là

\(\begin{array}{l}f'\left( x \right) = {\left[ {{{\log }_2}\left( {{2^x} + \sqrt {{4^x} + 1} } \right)} \right]^\prime } = \frac{{{{\left( {{2^x} + \sqrt {{4^x} + 1} } \right)}^\prime }}}{{\left( {{2^x} + \sqrt {{4^x} + 1} } \right).\ln 2}}\\ = \frac{{{2^x}.\ln 2 + \frac{{{4^x}.\ln 4}}{{2\sqrt {{4^x} + 1} }}}}{{\left( {{2^x} + \sqrt {{4^x} + 1} } \right).\ln 2}} = \frac{{{2^x} + \frac{{{4^x}}}{{\sqrt {{4^x} + 1} }}}}{{\left( {{2^x} + \sqrt {{4^x} + 1} } \right)}} = \frac{{{2^x}\left( {{2^x} + \sqrt {{4^x} + 1} } \right)}}{{\sqrt {{4^x} + 1} \left( {{2^x} + \sqrt {{4^x} + 1} } \right)}} = \frac{{{2^x}}}{{\sqrt {{4^x} + 1} }}.\end{array}\)