Nếu \[f''\left( x \right) = \frac{{2\sin x}}{{{{\cos }^3}x}}\], thì f(x) bằng:

A.\[12\,m/{s^2}\]

B. \[8\,m/{s^2}\]

C. \[7\,m/{s^2}\]

D. \[6\,m/{s^2}\]

A.\[{y^{\left( n \right)}} = \frac{{{2^n}.{a^n}.n!}}{{{{\left( {ax + b} \right)}^{n + 1}}}}\]

B. \[{y^{\left( n \right)}} = \frac{{{{\left( { - 1} \right)}^n}{a^n}n!}}{{{{\left( {x + 1} \right)}^{n + 1}}}}\]

C. \[{y^{\left( n \right)}} = \frac{{{{\left( { - 1} \right)}^n}.n!}}{{{{\left( {ax + b} \right)}^{n + 1}}}}\]

D. \[{y^{\left( n \right)}} = \frac{{{{\left( { - 1} \right)}^n}.{a^n}.n!}}{{{{\left( {ax + b} \right)}^{n + 1}}}}\]

A.\[{y^{\left( 4 \right)}} = \frac{{7.4!}}{{{{\left( {x - 3} \right)}^5}}} - \frac{{5.4!}}{{{{\left( {x - 2} \right)}^5}}}\]

B. \[{y^{\left( 4 \right)}} = \frac{{5.4!}}{{{{\left( {x - 3} \right)}^5}}} - \frac{{2.4!}}{{{{\left( {x - 2} \right)}^5}}}\]

C. \[{y^{\left( 4 \right)}} = \frac{{5.4!}}{{{{\left( {x - 2} \right)}^5}}} - \frac{{7.4!}}{{{{\left( {x - 3} \right)}^5}}}\]

D. \[{y^{\left( 4 \right)}} = \frac{7}{{{{\left( {x - 3} \right)}^4}}} - \frac{5}{{{{\left( {x - 2} \right)}^4}}}\]

A.−cosx 

B.sinx 

C.−sinx 

D.cosx 

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

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

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

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

A.\[dy = {\left( {{x^2} + 2x} \right)^\prime }dx\]

B. \[dx = {\left( {{x^2} + 2x} \right)^\prime }dy\]

C. \[dy = \left( {{x^2} + 2x} \right)dx\]

D. \[dy = \frac{1}{{{x^2} + 2x}}dx\]

A.\[x \in \left( {1; + \infty } \right).\]

B. \[x \in \left( { - \infty ;1} \right) \setminus \left\{ 0 \right\}.\]

C. \[x \in \left( { - 1;1} \right).\]

D. \[x \in \left( { - 2;2} \right).\]