Cho hàm số \(f(x) = \dfrac{{{x^3}}}{{x - 1}}\). Phương trình \({f^\prime }(x) = 0\) có tập nghiệm \(S\) là

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\left\{ { - 1;2} \right\}\).

\(\left\{ { - 1;3} \right\}\).

\(\left\{ {0;4} \right\}\).

\(\left\{ {1;2} \right\}\).

\(y'\left( 1 \right) =  - 4\).

\(y'\left( 1 \right) =  - 5\).

\(y'\left( 1 \right) =  - 3\).

\(y'\left( 1 \right) =  - 2\).