Hàm số \[y = {\left( {{x^2} + {\rm{ }}1} \right)^3}\] có đạo hàm cấp ba là:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[4y - y' = 0\]

\[4y + y'' = 0\]

\(y = y'tan2x\)

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

Chỉ \[\left( I \right)\] đúng

Chỉ \[\left( {II} \right)\] đúng

Cả hai đều đúng

Cả hai đều sai

\[0\]

\[ - 1\]

\( - 2\)

\(5\)

\[{y^{\left( {10} \right)}}\left( 1 \right) = 0\]

\[{y^{\left( {10} \right)}}\left( 1 \right) = 10a + b\]

\[{y^{\left( {10} \right)}}\left( 1 \right) = 5a\]

\[{y^{\left( {10} \right)}}\left( 1 \right) = 10a\]