Bài tập 2.33 trang 117 SBT Toán 12

a) \({\left[ {{{\log }_8}\left( {{x^2} - 3x - 4} \right)} \right]^\prime } = \frac{{2x - 3}}{{\left( {{x^2} - 3x - 4} \right)\ln 8}}\)

b) \({\left[ {{{\log }_{\sqrt 3 }}\left( { - {x^2} + 5x + 6} \right)} \right]^\prime } = \frac{{ - 2x + 5}}{{\left( { - {x^2} + 5x + 6} \right)\ln \sqrt 3 }}\)

c)

\({\left( {{{\log }_{0,7}}\frac{{{x^2} - 9}}{{x + 5}}} \right)^\prime } = \frac{{\frac{{2x\left( {x + 5} \right) - \left( {{x^2} - 9} \right)}}{{{{\left( {x + 5} \right)}^2}}}}}{{\left( {\frac{{{x^2} - 9}}{{x + 5}}} \right)\ln \left( {0,7} \right)}} = \frac{{{x^2} + 10x + 9}}{{\left( {x + 5} \right)\left( {{x^2} - 9} \right)\ln \left( {0,7} \right)}}\)

d) \({\left( {{{\log }_{\frac{1}{3}}}\frac{{x - 4}}{{x + 4}}} \right)^\prime } = \frac{{\frac{8}{{{{\left( {x + 4} \right)}^2}}}}}{{\frac{{x - 4}}{{x + 4}}\ln \frac{1}{3}}} = \frac{8}{{\left( {{x^2} - 16} \right)\ln 3}}\)

e) \({\left[ {{{\log }_\pi }\left( {{2^x} - 2} \right)} \right]^\prime } = \frac{{{2^x}\ln 2}}{{\left( {{2^x} - 2} \right)\ln \pi }}\)

f) \({\left[ {{{\log }_3}\left( {{3^{x - 1}} - 9} \right)} \right]^\prime } = \frac{{{3^{x - 1}}\ln 3}}{{\left( {{3^{x - 1}} - 9} \right)\ln 3}} = \frac{{{3^{x - 1}}}}{{{3^{x - 1}} - 9}}\)

-- Mod Toán 12 HỌC247