Đạo hàm cấp hai của hàm số $y = 3{x^2} - 2021x + 2020$ là

$dy = \left( {{x^2} + 2x} \right)'dx$ 

$dx = \left( {{x^2} + 2x} \right)'dy$ 

$dy = \left( {{x^2} + 2x} \right)dx$ 

$dy = \dfrac{1}{{{x^2} + 2x}}dx$

$y'' = 0$

$y'' = \dfrac{1}{{{{\left( {x - 2} \right)}^2}}}$

$y'' =  - \dfrac{4}{{{{\left( {x - 2} \right)}^3}}}$

$y'' = \dfrac{4}{{{{\left( {x - 2} \right)}^3}}}$

$y''' = 12x\left( {{x^2} + 1} \right)$ 

$y''' = 24x\left( {{x^2} + 1} \right)$ 

$y''' = 24x\left( {5{x^2} + 3} \right)$ 

$y''' =  - 12x\left( {{x^2} + 1} \right)$ 

$y'' = \dfrac{1}{{\left( {2x + 5} \right)\sqrt {2x + 5} }}$ 

$y'' = \dfrac{1}{{\sqrt {2x + 5} }}$ 

$y'' =  - \dfrac{1}{{\left( {2x + 5} \right)\sqrt {2x + 5} }}$ 

$y''=-\dfrac{1}{\sqrt{2x+5}}$

$y'' =  - \dfrac{{2\sin x}}{{{{\cos }^3}x}}$ 

$y'' = \dfrac{1}{{{{\cos }^2}x}}$ 

$y'' =  - \dfrac{1}{{{{\cos }^2}x}}$ 

$y'' = \dfrac{{2\sin x}}{{{{\cos }^3}x}}$ 

$M = 0.$

$M = 20.$

$M = 40.$

$M = 100.$

$\left[ { - 1;2} \right]$ 

$\left( { - \infty ;0} \right]$ 

$\left\{ { - 1} \right\}$ 

$\emptyset$