Cho hàm số $f\left(x\right)=\left\{\begin{array}{l}{x}^2-x khi x\ge 0\\ 2x khi x<0\end{array}\right.$. Kết quả nào dưới đây đúng?

$f^{\text{'}}\left(0^{+}\right)=1$

$f^{\text{'}}\left(0^{-}\right)=1$

$df\left(0\right)=dx$

Hàm số không có vi phân tại x=0

$dy=\frac{-4x}{{(1+x^2)}^2}dx$

$dy=\frac{-4}{{(1+x^2)}^2}dx$

$dy=\frac{-4}{1+x^2}dx$

$dy=\frac{-dx}{{(1+x^2)}^2}$

\(y'' = 0\)

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

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

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

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

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

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

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

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

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

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

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

\({y^{(5)}} = - \frac{{120}}{{{{(x + 1)}^6}}}\)

\({y^{(5)}} = \frac{{120}}{{{{(x + 1)}^6}}}\)

\({y^{(5)}} = \frac{1}{{{{(x + 1)}^6}}}\)

\({y^{(5)}} = - \frac{1}{{{{(x + 1)}^6}}}\)

\[{y^{\left( 5 \right)}} = - \frac{{120}}{{{{\left( {x + 1} \right)}^6}}}\]

\[{y^{\left( 5 \right)}} = \frac{{120}}{{{{\left( {x + 1} \right)}^5}}}\]

\[{y^{\left( 5 \right)}} = \frac{1}{{{{\left( {x + 1} \right)}^5}}}\]

\[{y^{\left( 5 \right)}} = - \frac{1}{{{{\left( {x + 1} \right)}^5}}}\]

\[y'' = - \frac{{2\sin x}}{{{{\cos }^3}x}}\]

\[y'' = \frac{1}{{{{\cos }^2}x}}\]

\[y'' = - \frac{1}{{{{\cos }^2}x}}\]

\[y'' = \frac{{2\sin x}}{{{{\cos }^3}x}}\]

\[y' = \sin \left( {x + \frac{\pi }{2}} \right)\]

\[y'' = \sin \left( {x + \pi } \right)\]

\[y''' = \sin \left( {x + \frac{{3\pi }}{2}} \right)\]

\[{y^{\left( 4 \right)}} = \sin \left( {2\pi - x} \right)\]

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

\[y'' = \frac{2}{{{{\left( {1 - x} \right)}^3}}}\]

\[y'' = \frac{{ - 2}}{{{{\left( {1 - x} \right)}^3}}}\]

\[y'' = \frac{2}{{{{\left( {1 - x} \right)}^4}}}\]