Tính đạo hàm cấp hai của các hàm số sau:

a) y = ln|2x – 1|;                                   

b) $y=\tan \left(x+\frac{\pi }{3}\right)$

a) y' = (ln|2x – 1|)' = $\frac{{\left(2x-1\right)}^{\text{'}}}{2x-1}$$=\frac{2}{2x-1}$

$y^{\text{'}\text{'}}={\left(\frac{2}{2x-1}\right)}^{\text{'}}$$=2\cdot \frac{-2}{{\left(2x-1\right)}^2}$$=\frac{-4}{{\left(2x-1\right)}^2}$

Vậy $y^{\text{'}\text{'}}=\frac{-4}{{\left(2x-1\right)}^2}$

b) $y^{\text{'}}={\left[\tan \left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$$=\frac{{\left(x+\frac{\pi }{3}\right)}^{\text{'}}}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$=\frac{1}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$1+{\tan }^2\left(x+\frac{\pi }{3}\right)$

$y^{\text{'}\text{'}}={\left[1+{\tan }^2\left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$$=2\tan \left(x+\frac{\pi }{3}\right)\cdot {\left[\tan \left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$

$=2\tan \left(x+\frac{\pi }{3}\right)\cdot \frac{{\left(x+\frac{\pi }{3}\right)}^{\text{'}}}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$=2\tan \left(x+\frac{\pi }{3}\right)\cdot \frac{1}{co{s}^2\left(x+\frac{\pi }{3}\right)}$

$=2\tan \left(x+\frac{\pi }{3}\right)\cdot \left(1+{\tan }^2\left(x+\frac{\pi }{3}\right)\right)$$=2\tan \left(x+\frac{\pi }{3}\right)+2{\tan }^3\left(x+\frac{\pi }{3}\right)$

Vậy $y^{\text{'}\text{'}}=2\tan \left(x+\frac{\pi }{3}\right)+2{\tan }^3\left(x+\frac{\pi }{3}\right)$