Bài 5 (Trang 78 SGK Toán Giải Tích 12):

Tính đạo hàm của các hàm số

a) y = 3x² - ln x + 4.sin x;

b) y = log(x²+x+1);

c) $\mathrm{y} = \frac{{\log }_3\mathrm{x}}{\mathrm{x}}$.

Hướng dẫn giải:

a) $\mathrm{y}\text{'} = (3\mathrm{x}2 - \ln \mathrm{x} + 4.\sin \mathrm{x})\text{'} = \left(3\mathrm{x}2\right)\text{'} - (\ln \mathrm{x})\text{'} + (4.\sin \mathrm{x})\text{'} = 6\mathrm{x} - \frac{1}{\mathrm{x}} + 4.\cos \mathrm{x}$.

b) $\mathrm{y}\text{'} = \left[\log \left({\mathrm{x}}^2+\mathrm{x}+1\right)\right]\text{'} = \frac{\left({\mathrm{x}}^2+\mathrm{x}+1\right)\text{'}}{\left({\mathrm{x}}^2+\mathrm{x}+1\right). \ln 10} = \frac{2\mathrm{x}+1}{\left({\mathrm{x}}^2+\mathrm{x}+1\right).\ln 10}$.

c) $\mathrm{y}\text{'} = {\left(\frac{{\log }_3\mathrm{x}}{\mathrm{x}}\right)}^{\text{'}} = \frac{{\left({\log }_3\mathrm{x}\right)}^{\text{'}}.\mathrm{x}-\left(\mathrm{x}\right)\text{'}.{\log }_3\mathrm{x}}{{\mathrm{x}}^2} = \frac{{\displaystyle \frac{1}{\mathrm{x}.\ln 3}}.\mathrm{x} - {\log }_3\mathrm{x}}{{\mathrm{x}}^2} = \frac{{\displaystyle \frac{1}{\ln 3}}-{\log }_3\mathrm{x}}{{\mathrm{x}}^2} = \frac{1-\ln 3.{\log }_3\mathrm{x}}{{\mathrm{x}}^2.\ln 3} = \frac{1-\ln \mathrm{x}}{{\mathrm{x}}^2 - \ln 3}$.