Tìm $m$ để hàm số $y = \dfrac{{m{x^3}}}{3} - m{x^2} + \left( {3m - 1} \right)x + 1$ có $y' \le 0\,\,\forall x \in R$

${(u + v)^\prime } = {u^\prime } + {v^\prime }$.

${(u - v)^\prime } = {u^\prime } - {v^\prime }$.

${(ku)^\prime } = k{u^\prime }$ ( $k$ là hằng số).

${(uv)^\prime } = {u^\prime }{v^\prime }$.

${y^\prime } = \dfrac{{2x}}{{\sqrt {{x^2} + 1} }}$.

${y^\prime } = \dfrac{x}{{\sqrt {{x^2} + 1} }}$.

${y^\prime } = \dfrac{{2x + 1}}{{2\sqrt {{x^2} + 1} }}$.

${y^\prime } = \dfrac{1}{{2\sqrt {{x^2} + 1} }}$.

${y^\prime }\left( {\dfrac{\pi }{6}} \right) = \sqrt 3$.

${y^\prime }\left( {\dfrac{\pi }{6}} \right) =  - 1$.

${y^\prime }\left( {\dfrac{\pi }{6}} \right) = 1$.

${y^\prime }\left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{2}$.

$f'\left( { - 1} \right) = 4.$

$f'\left( { - 1} \right) = 14.$

$f'\left( { - 1} \right) = 15.$

$f'\left( { - 1} \right) = 24.$

$y' = 5{\left( {1 - {x^3}} \right)^4}$ .

$y' =  - 15{x^2}{\left( {1 - {x^3}} \right)^4}$.

$y' =  - 3{\left( {1 - {x^3}} \right)^4}$.

$y' =  - 5{x^2}{\left( {1 - {x^3}} \right)^4}$.

$\left\{ { - 1;2} \right\}$.

$\left\{ { - 1;3} \right\}$.

$\left\{ {0;4} \right\}$.

$\left\{ {1;2} \right\}$.

$y'\left( 1 \right) =  - 4$.

$y'\left( 1 \right) =  - 5$.

$y'\left( 1 \right) =  - 3$.

$y'\left( 1 \right) =  - 2$.