Bài 2.66 trang 133 SBT giải tích 12

Đề bài

Tính đạo hàm của các hàm số sau:

a) \(\displaystyle y = \frac{1}{{{{(2 + 3x)}^2}}}\)

b) \(\displaystyle y = \sqrt[3]{{{{(3x - 2)}^2}}}\left( {x \ne \frac{2}{3}} \right)\)

c) \(\displaystyle y = \frac{1}{{\sqrt[3]{{3x - 7}}}}\)    d) \(\displaystyle y = 3{x^{ - 3}} - {\log _3}x\)

e) \(\displaystyle y = (3{x^2} - 2){\log _2}x\)   g) \(\displaystyle y = \ln (\cos x)\)

h) \(\displaystyle y = {e^x}\sin x\)        i) \(\displaystyle y = \frac{{{e^x} - {e^{ - x}}}}{x}\)

Lời giải chi tiết

a) \(\displaystyle y = \frac{1}{{{{(2 + 3x)}^2}}} = {\left( {2 + 3x} \right)^{ - 2}}\)\(\displaystyle  \Rightarrow y' =  - 2\left( {2 + 3x} \right)'{\left( {2 + 3x} \right)^{ - 3}}\)\(\displaystyle  =  - 6{(2 + 3x)^{ - 3}}\)

b) Với \(\displaystyle x > \frac{2}{3}\) thì \(\displaystyle y = {\left( {3x - 2} \right)^{\frac{2}{3}}}\) nên \(\displaystyle y' = 2{(3x - 2)^{ - \frac{1}{3}}} = \frac{2}{{\sqrt[3]{{3x - 2}}}}\).

Với \(\displaystyle x < \frac{2}{3}\) thì \(\displaystyle y =  - {\left( {2 - 3x} \right)^{\frac{2}{3}}}\) nên \(\displaystyle y' =  - 2{\left( {2 - 3x} \right)^{ - \frac{1}{3}}}\) \(\displaystyle  = \frac{{ - 2}}{{\sqrt[3]{{2 - 3x}}}} = \frac{2}{{\sqrt[3]{{3x - 2}}}}\).

Vậy \(\displaystyle y' = \frac{2}{{\sqrt[3]{{3x - 2}}}}\left( {x \ne \frac{2}{3}} \right)\).

c) Với \(\displaystyle x > \frac{7}{3}\) thì \(\displaystyle y = \frac{1}{{\sqrt[3]{{3x - 7}}}} = {\left( {3x - 7} \right)^{ - \frac{1}{3}}}\) nên \(\displaystyle y' =  - \frac{1}{3}.3{\left( {3x - 7} \right)^{ - \frac{4}{3}}}\) \(\displaystyle  =  - {\left( {3x - 7} \right)^{ - \frac{4}{3}}} =  - \frac{1}{{\sqrt[3]{{{{\left( {3x - 7} \right)}^4}}}}}\)

Với \(\displaystyle x < \frac{7}{3}\) thì \(\displaystyle y = \frac{1}{{\sqrt[3]{{3x - 7}}}} =  - {\left( {7 - 3x} \right)^{ - \frac{1}{3}}}\) nên:

\(\displaystyle y' = \frac{1}{3}.\left( { - 3} \right){\left( {7 - 3x} \right)^{ - \frac{4}{3}}}\) \(\displaystyle  =  - {\left( {7 - 3x} \right)^{ - \frac{4}{3}}} =  - \frac{1}{{\sqrt[3]{{{{\left( {7 - 3x} \right)}^4}}}}}\)\(\displaystyle  =  - \frac{1}{{\sqrt[3]{{{{\left( {3x - 7} \right)}^4}}}}}\)

Vậy \(\displaystyle y' =  - \frac{1}{{\sqrt[3]{{{{(3x - 7)}^4}}}}}\)

d) \(\displaystyle y = 3{x^{ - 3}} - {\log _3}x\) \(\displaystyle  \Rightarrow y' = 3.\left( { - 3} \right).{x^{ - 4}} - \frac{1}{{x\ln 3}}\) \(\displaystyle  =  - 9{x^{ - 4}} - \frac{1}{{x\ln 3}}\)

e) \(\displaystyle y = (3{x^2} - 2){\log _2}x\)\(\displaystyle  \Rightarrow y' = 6x{\log _2}x + \left( {3{x^2} - 2} \right).\frac{1}{{x\ln 2}}\) \(\displaystyle  = 6x{\log _2}x + \frac{{3{x^2} - 2}}{{x\ln 2}}\)

g) \(\displaystyle y = \ln (\cos x)\)\(\displaystyle  \Rightarrow y' = \frac{{\left( {\cos x} \right)'}}{{\cos x}}\) \(\displaystyle  =  - \frac{{\sin x}}{{\cos x}} =  - \tan x\)

h) \(\displaystyle y = {e^x}\sin x\)\(\displaystyle  \Rightarrow y' = {e^x}\sin x + {e^x}\cos x\) \(\displaystyle  = {e^x}(\sin x + \cos x)\)

i) \(\displaystyle y = \frac{{{e^x} - {e^{ - x}}}}{x}\)\(\displaystyle  \Rightarrow y' = \frac{{\left( {{e^x} + {e^{ - x}}} \right)x - \left( {{e^x} - {e^{ - x}}} \right)}}{{{x^2}}}\) \(\displaystyle  = \frac{{x({e^x} + {e^{ - x}}) - {e^x} + {e^{ - x}}}}{{{x^2}}}\)