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

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

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

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

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

\(y'' = 0\)

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

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

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

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

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

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

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

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

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

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

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

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

\(y'' = \dfrac{1}{{{{\cos }^2}x}}\) 

\(y'' =  - \dfrac{1}{{{{\cos }^2}x}}\) 

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

\(M = 0.\)

\(M = 20.\)

\(M = 40.\)

\(M = 100.\)

\(\left[ { - 1;2} \right]\) 

\(\left( { - \infty ;0} \right]\) 

\(\left\{ { - 1} \right\}\) 

\(\emptyset \)