Nếu f ′′ ( x ) = 2 sin x/cos^3 x , thì f(x) bằng:

14/22

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

\[\frac{1}{{\cos x}}\]

\[ - \frac{1}{{\cos x}}\]

\[\cot x\]

\[\tan x\]

Giải thích

\[\begin{array}{*{20}{l}}{y = \frac{1}{{\cos x}}}\\{y' = \frac{{ - {{\left( {\cos x} \right)}^\prime }}}{{{{\cos }^2}x}} = \frac{{\sin x}}{{{{\cos }^2}x}}}\\{y'' = \frac{{\cos x.{{\cos }^2}x - \sin x.2\cos x{{\left( {\cos x} \right)}^\prime }}}{{{{\left( {{{\cos }^2}x} \right)}^2}}} = \frac{{{{\cos }^3}x + 2{{\sin }^2}x\cos x}}{{{{\cos }^4}x}} = \frac{{{{\cos }^2}x + 2{{\sin }^2}x}}{{{{\cos }^3}x}}}\end{array}\]

Đáp án B:

\[\begin{array}{*{20}{l}}{y = - \frac{1}{{\cos x}}}\\{y' = \frac{{{{\left( {\cos x} \right)}^\prime }}}{{{{\cos }^2}x}} = - \frac{{\sin x}}{{{{\cos }^2}x}}}\\{y'' = - \frac{{\cos x.{{\cos }^2}x - \sin x.2\cos x{{\left( {\cos x} \right)}^\prime }}}{{{{\cos }^4}x}} = \frac{{ - {{\cos }^3}x - 2{{\sin }^2}x\cos x}}{{{{\cos }^4}x}} = - \frac{{{{\cos }^2}x + 2{{\sin }^2}x}}{{{{\cos }^4}x}}}\end{array}\]

Đáp án C:

\[\begin{array}{*{20}{l}}{y = \cot x}\\{y' = - \frac{1}{{{{\sin }^2}x}}}\\{y' = \frac{{2\sin x{{\left( {\sin x} \right)}^\prime }}}{{{{\sin }^4}x}} = \frac{{2\sin x\cos x}}{{{{\sin }^4}x}} = \frac{{2\cos x}}{{{{\sin }^3}x}}}\end{array}\]

Đáp án D:

\[\begin{array}{*{20}{l}}{y = \tan x}\\{y' = \frac{1}{{{{\cos }^2}x}}}\\{y'' = \frac{{ - 2\cos x{{\left( {\cos x} \right)}^\prime }}}{{{{\cos }^4}x}} = \frac{{2\sin x\cos x}}{{{{\cos }^4}x}} = \frac{{2\sin x}}{{{{\cos }^3}x}}}\end{array}\]

Đáp án cần chọn là: D