Hàm số \(y = \frac{{{x^2} + x + 1}}{{x + 1}}\) có đạo hàm cấp \(5\) bằng :

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

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

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

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

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

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

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

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

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

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

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

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

\[4y - y' = 0\]

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

\(y = y'tan2x\)

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

Chỉ \[\left( I \right)\] đúng

Chỉ \[\left( {II} \right)\] đúng

Cả hai đều đúng

Cả hai đều sai

\[0\]

\[ - 1\]

\( - 2\)

\(5\)

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

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

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

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

4 và 16

5 và 17

6 và 18

7 và 19

\({y^{(n)}} = \frac{{{{(1)}^{n - 1}}.3.n!}}{{{{(x + 2)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^{n - 1}}.n!}}{{{{(x + 2)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^{n - 1}}.3.n!}}{{{{(x - 2)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^{n - 1}}.3.n!}}{{{{(x + 2)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{(2)}^n}.7.n!}}{{{{(x - 2)}^{n + 1}}}} - \frac{{{{(1)}^n}.5.n!}}{{{{(x - 3)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^{n + 1}}.7.n!}}{{{{(x - 2)}^{n + 1}}}} - \frac{{{{( - 1)}^{n + 1}}.5.n!}}{{{{(x - 3)}^{n + 1}}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^n}.7.n!}}{{{{(x - 2)}^n}}} - \frac{{{{( - 1)}^n}.5.n!}}{{{{(x - 3)}^n}}}\)

\({y^{(n)}} = \frac{{{{( - 1)}^n}.7.n!}}{{{{(x - 2)}^{n + 1}}}} - \frac{{{{( - 1)}^n}.5.n!}}{{{{(x - 3)}^{n + 1}}}}\)