Google
Câu hỏi: Tính đạo hàm của các hàm số đã cho ở bài tập 2.32.
a) \(\displaystyle y = {\log _8}({x^2} - 3x - 4)\)
b) \(\displaystyle y = {\log _{\sqrt 3 }}( - {x^2} + 5x + 6)\)
c) \(\displaystyle y = {\log _{0,7}}\frac {{{x^2} - 9}}{{x + 5}}\)
d) \(\displaystyle y = {\log _{\frac {1}{3}}}\frac {{x - 4}}{{x + 4}}\)
e) \(\displaystyle y = {\log _\pi }({2^x} - 2)\)
g) \(\displaystyle y = {\log _3}({3^{x - 1}} - 9)\)
a) \(\displaystyle y = {\log _8}({x^2} - 3x - 4)\)
b) \(\displaystyle y = {\log _{\sqrt 3 }}( - {x^2} + 5x + 6)\)
c) \(\displaystyle y = {\log _{0,7}}\frac {{{x^2} - 9}}{{x + 5}}\)
d) \(\displaystyle y = {\log _{\frac {1}{3}}}\frac {{x - 4}}{{x + 4}}\)
e) \(\displaystyle y = {\log _\pi }({2^x} - 2)\)
g) \(\displaystyle y = {\log _3}({3^{x - 1}} - 9)\)
Phương pháp giải
Sử dụng công thức đạo hàm \(\displaystyle \left( {{{\log }_a}u} \right)' = \frac {{u'}}{{u\ln a}}\).
Lời giải chi tiết
a) \(\displaystyle y' = \frac {{\left( {{x^2} - 3x - 4} \right)'}}{{\left({{x^2} - 3x - 4} \right)\ln 8}}\)\(\displaystyle = \frac {{2x - 3}}{{({x^2} - 3x - 4)\ln 8}}\)
b) \(\displaystyle y' = \frac {{\left( { - {x^2} + 5x + 6} \right)'}}{{\left({ - {x^2} + 5x + 6} \right)\ln \sqrt 3 }}\)\(\displaystyle = \frac {{ - 2x + 5}}{{\left( { - {x^2} + 5x + 6} \right).\frac {1}{2}\ln 3}}\) \(\displaystyle = \frac {{ - 4x + 10}}{{( - {x^2} + 5x + 6)\ln 3}}\)
c) \(\displaystyle y' = \frac {{\left( {\frac {{{x^2} - 9}}{{x + 5}}} \right)'}}{{\left({\frac {{{x^2} - 9}}{{x + 5}}} \right)\ln 0,7}}\)\(\displaystyle = \frac {{2x\left( {x + 5} \right) - \left({{x^2} - 9} \right)}}{{{{\left({x + 5} \right)}^2}}}.\frac {{x + 5}}{{{x^2} - 9}}.\frac {1}{{\ln 0,7}}\) \(\displaystyle = \frac {{{x^2} + 10x + 9}}{{({x^2} - 9)(x + 5)\ln 0,7}}\)
d) \(\displaystyle y' = \frac {{\left( {\frac {{x - 4}}{{x + 4}}} \right)'}}{{\frac {{x - 4}}{{x + 4}}\ln \frac {1}{3}}}\)\(\displaystyle = \frac {8}{{{{\left( {x + 4} \right)}^2}}}.\frac {{x + 4}}{{x - 4}}.\frac {1}{{ - \ln 3}}\) \(\displaystyle = \frac {8}{{ - \left( {x + 4} \right)\left({x - 4} \right).\ln 3}}\) \(\displaystyle = \frac {8}{{(16 - {x^2})\ln 3}}\)
e) \(\displaystyle y' = \frac {{\left( {{2^x} - 2} \right)'}}{{\left({{2^x} - 2} \right)\ln \pi }}\)\(\displaystyle = \frac {{{2^x}\ln 2}}{{\left( {{2^x} - 2} \right)\ln \pi }}\)
g) \(\displaystyle y' = \frac {{\left( {{3^{x - 1}} - 9} \right)'}}{{\left({{3^{x - 1}} - 9} \right)\ln 3}}\)\(\displaystyle = \frac {{{3^{x - 1}}\ln 3}}{{\left( {{3^{x - 1}} - 9} \right)\ln 3}} = \frac {{{3^{x - 1}}}}{{{3^{x - 1}} - 9}}\)
Sử dụng công thức đạo hàm \(\displaystyle \left( {{{\log }_a}u} \right)' = \frac {{u'}}{{u\ln a}}\).
Lời giải chi tiết
a) \(\displaystyle y' = \frac {{\left( {{x^2} - 3x - 4} \right)'}}{{\left({{x^2} - 3x - 4} \right)\ln 8}}\)\(\displaystyle = \frac {{2x - 3}}{{({x^2} - 3x - 4)\ln 8}}\)
b) \(\displaystyle y' = \frac {{\left( { - {x^2} + 5x + 6} \right)'}}{{\left({ - {x^2} + 5x + 6} \right)\ln \sqrt 3 }}\)\(\displaystyle = \frac {{ - 2x + 5}}{{\left( { - {x^2} + 5x + 6} \right).\frac {1}{2}\ln 3}}\) \(\displaystyle = \frac {{ - 4x + 10}}{{( - {x^2} + 5x + 6)\ln 3}}\)
c) \(\displaystyle y' = \frac {{\left( {\frac {{{x^2} - 9}}{{x + 5}}} \right)'}}{{\left({\frac {{{x^2} - 9}}{{x + 5}}} \right)\ln 0,7}}\)\(\displaystyle = \frac {{2x\left( {x + 5} \right) - \left({{x^2} - 9} \right)}}{{{{\left({x + 5} \right)}^2}}}.\frac {{x + 5}}{{{x^2} - 9}}.\frac {1}{{\ln 0,7}}\) \(\displaystyle = \frac {{{x^2} + 10x + 9}}{{({x^2} - 9)(x + 5)\ln 0,7}}\)
d) \(\displaystyle y' = \frac {{\left( {\frac {{x - 4}}{{x + 4}}} \right)'}}{{\frac {{x - 4}}{{x + 4}}\ln \frac {1}{3}}}\)\(\displaystyle = \frac {8}{{{{\left( {x + 4} \right)}^2}}}.\frac {{x + 4}}{{x - 4}}.\frac {1}{{ - \ln 3}}\) \(\displaystyle = \frac {8}{{ - \left( {x + 4} \right)\left({x - 4} \right).\ln 3}}\) \(\displaystyle = \frac {8}{{(16 - {x^2})\ln 3}}\)
e) \(\displaystyle y' = \frac {{\left( {{2^x} - 2} \right)'}}{{\left({{2^x} - 2} \right)\ln \pi }}\)\(\displaystyle = \frac {{{2^x}\ln 2}}{{\left( {{2^x} - 2} \right)\ln \pi }}\)
g) \(\displaystyle y' = \frac {{\left( {{3^{x - 1}} - 9} \right)'}}{{\left({{3^{x - 1}} - 9} \right)\ln 3}}\)\(\displaystyle = \frac {{{3^{x - 1}}\ln 3}}{{\left( {{3^{x - 1}} - 9} \right)\ln 3}} = \frac {{{3^{x - 1}}}}{{{3^{x - 1}} - 9}}\)