Bài 2.33 trang 117 SBT giải tích 12

Đề bài

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) \(\displaystyle y = {\log _3}({3^{x – 1}} – 9)\)

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) \(\displaystyle y’ = \frac {{\left( {{3^{x – 1}} – 9} \right)’}}{{\left( {{3^{x – 1}} – 9} \right)\ln 3}}\)\(\displaystyle  = \frac {{{3^{x – 1}}\ln 3}}{{\left( {{3^{x – 1}} – 9} \right)\ln 3}} = \frac {{{3^{x – 1}}}}{{{3^{x – 1}} – 9}}\)

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