Nếu \(f''\left( x \right) = \frac{{2\sin x}}{{{{\cos }^3}x}}\) thì f(x) bằng:
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Lời giải:
Báo sai\(\begin{array}{l} {\left( {\tan x} \right)^\prime } = \frac{1}{{{{\cos }^2}x}} \Rightarrow {\left( {\tan x} \right)^\prime }^\prime = \frac{{2{\mathop{\rm sinx}\nolimits} }}{{{{\cos }^3}x}} \Rightarrow {\rm{A}}\,{\rm{True}}\\ {\left( {\cot x} \right)^\prime } = - \frac{1}{{si{n^2}x}} \Rightarrow {\left( {\cot x} \right)^\prime }^\prime = \frac{{ - 2{\mathop{\rm cosx}\nolimits} }}{{{{\cos }^3}x}} \Rightarrow {\rm{B}}\,\,{\rm{False}}\\ {\left( { - \frac{1}{{\cos x}}} \right)^\prime } = \frac{{{\mathop{\rm sinx}\nolimits} }}{{co{s^2}x}} \Rightarrow {\left( { - \frac{1}{{\cos x}}} \right)^\prime }^\prime = \frac{{co{s^2}x + 2{{\sin }^2}x}}{{co{s^3}x}} \Rightarrow {\rm{C}}\,\,{\rm{False}}\\ {\left( {\frac{1}{{{{\cos }^2}x}}} \right)^\prime } = \frac{{2{\mathop{\rm sinx}\nolimits} }}{{{{\cos }^3}x}} \Rightarrow {\left( {\frac{1}{{{{\cos }^2}x}}} \right)^\prime }^\prime = \frac{{2{{\cos }^2}x + 6{{\sin }^2}x}}{{{{\cos }^4}x}} \Rightarrow {\rm{D}}\,\,{\rm{False}} \end{array}\)
