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

Đáp án A:

$\begin{array}{l}y = \dfrac{1}{{\cos x}}\\y' = \dfrac{{ - \left( {\cos x} \right)'}}{{{{\cos }^2}x}} = \dfrac{{\sin x}}{{{{\cos }^2}x}}\\y'' = \dfrac{{\cos x.{{\cos }^2}x - \sin x.2\cos x\left( {\cos x} \right)'}}{{{{\left( {{{\cos }^2}x} \right)}^2}}} = \dfrac{{{{\cos }^3}x + 2{{\sin }^2}x\cos x}}{{{{\cos }^4}x}} = \dfrac{{{{\cos }^2}x + 2{{\sin }^2}x}}{{{{\cos }^3}x}}\end{array}$. 

Đáp án B:

$\begin{array}{l}y =  - \dfrac{1}{{\cos x}}\\y' = \dfrac{{\left( {\cos x} \right)'}}{{{{\cos }^2}x}} =  - \dfrac{{\sin x}}{{{{\cos }^2}x}}\\y'' =  - \dfrac{{\cos x.{{\cos }^2}x - \sin x.2\cos x\left( {\cos x} \right)'}}{{{{\cos }^4}x}} = \dfrac{{ - {{\cos }^3}x - 2{{\sin }^2}x\cos x}}{{{{\cos }^4}x}} =  - \dfrac{{{{\cos }^2}x + 2{{\sin }^2}x}}{{{{\cos }^4}x}}\end{array}$

Đáp án C:

$\begin{array}{l}y = \cot x\\y' =  - \dfrac{1}{{{{\sin }^2}x}}\\y' = \dfrac{{2\sin x\left( {\sin x} \right)'}}{{{{\sin }^4}x}} = \dfrac{{2\sin x\cos x}}{{{{\sin }^4}x}} = \dfrac{{2\cos x}}{{{{\sin }^3}x}}\end{array}$

Đáp án D:

$\begin{array}{l}y = \tan x\\y' = \dfrac{1}{{{{\cos }^2}x}}\\y'' = \dfrac{{ - 2\cos x\left( {\cos x} \right)'}}{{{{\cos }^4}x}} = \dfrac{{2\sin x\cos x}}{{{{\cos }^4}x}} = \dfrac{{2\sin x}}{{{{\cos }^3}x}}\end{array}$