Đề bài

Tìm đạo hàm cấp hai của mỗi hàm số sau: a) \(y = \frac{1}{{2x + 3}}\). b) \(y = {\log _3}x\). c) \(y = {2^x}\).

Lời giải chi tiết

a) \(y' = \left( {\frac{1}{{2x - 3}}} \right)' = \frac{{1'\left( {2x + 3} \right) - 1\left( {2x + 3} \right)'}}{{{{\left( {2x - 3} \right)}^2}}} =  - \frac{2}{{{{\left( {2x - 3} \right)}^2}}}\). \(y'' = \left[ { - \frac{2}{{{{\left( {2x - 3} \right)}^2}}}} \right]' =  - \frac{{2'{{\left( {2x - 3} \right)}^2} - 2\left[ {{{\left( {2x - 3} \right)}^2}} \right]'}}{{{{\left( {2x - 3} \right)}^4}}}\) \( =  - \frac{{ - 2.2\left( {2x - 3} \right)'\left( {2x - 3} \right)}}{{{{\left( {2x - 3} \right)}^4}}} = \frac{{8\left( {2x - 3} \right)}}{{{{\left( {2x - 3} \right)}^4}}} = \frac{8}{{{{\left( {2x - 3} \right)}^3}}}\). b) \(y' = \left( {{{\log }_3}x} \right)' = \frac{1}{{x\ln 3}}\). \( y'' = \left( {\frac{1}{{x\ln 3}}} \right)' =  - \frac{{\left( {x\ln 3} \right)'}}{{{{\left( {x\ln 3} \right)}^2}}} \) \(=  - \frac{{\ln 3}}{{{{\left( {x\ln 3} \right)}^2}}} =  - \frac{{\ln 3}}{{{{\left( {x\ln 3} \right)}^2}}} =  - \frac{1}{{x.\ln 3}}\). c) \(y' = \left( {{2^x}} \right)' = {2^x}.\ln 2\). \( y'' = \left( {{2^x}.\ln 2} \right)' = {2^x}.\ln 2.\ln 2 = {2^x}.{\left( {\ln 2} \right)^2}\).