Cho hàm số \(f\left( x \right) = {x^3} - 3{x^2} + 1\). Đạo hàm của hàm số f(x) âm khi và chỉ khi

\(y' = 4{x^3} - 6x + 3\) 

\(y' = 4{x^4} - 6x + 2\) 

\(y' = 4{x^3} - 3x + 2\)            

\(y' = 4{x^3} - 6x + 2\)

\( - \dfrac{3}{{{{\left( {x + 2} \right)}^2}}}\)             

\(\dfrac{3}{{x + 2}}\) 

\(\dfrac{3}{{{{\left( {x + 2} \right)}^2}}}\) 

\(\dfrac{2}{{{{\left( {x + 2} \right)}^2}}}\)

\(\dfrac{1}{6}\) 

\(\dfrac{1}{{12}}\) 

\( - \dfrac{1}{6}\) 

\( - \dfrac{1}{{12}}\)

$\{1\}$

\(\mathbb{R}\backslash \left\{ 1 \right\}\)

\(\emptyset \) 

$R$

\(y = \dfrac{{{x^3} + 1}}{x}\) 

\(y = \dfrac{{3\left( {{x^2} + x} \right)}}{{{x^3}}}\) 

\(y = \dfrac{{{x^3} + 5x - 1}}{x}\) 

\(y = \dfrac{{2{x^2} + x - 1}}{x}\) 

\( y'=- \dfrac{3}{{{x^4}}} + \dfrac{1}{{{x^3}}}\)

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

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

\(y'=\dfrac{3}{{{x^4}}} - \dfrac{1}{{{x^3}}}\)

\(\dfrac{a}{c}\) 

\(\dfrac{{ad - bc}}{{{{\left( {cx + d} \right)}^2}}}\) 

\(\dfrac{{ad + bc}}{{{{\left( {cx + d} \right)}^2}}}\)                         

\(\dfrac{{ad - bc}}{{cx + d}}\)