Tính đạo hàm cấp hai của các hàm số sau:

a) $y=\frac{x^4}{4}-2x^2+1$ ;                             

b) $y=\frac{2x+1}{x-1}$ .

a) Có $y^{\text{'}}={\left(\frac{x^4}{4}-2x^2+1\right)}^{\text{'}}$$=x^3-4x$ .

Có y" = (x³ – 4x)' = 3x² – 4.

Vậy y" = 3x² – 4.

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

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

$y^{\text{'}\text{'}}={\left[\frac{-3}{{\left(x-1\right)}^2}\right]}^{\text{'}}$$=-3{\left[\frac{1}{{\left(x-1\right)}^2}\right]}^{\text{'}}$$=-3\cdot \frac{-2}{{\left(x-1\right)}^3}$$=\frac{6}{{\left(x-1\right)}^3}$

Vậy $y^{\text{'}\text{'}}=\frac{6}{{\left(x-1\right)}^3}$