Hướng dẫn giải:

Chọn D.

\(f'\left( x \right) = \frac{2}{{\cos \left( {\pi x} \right)}} = 2.\left( {\cos \left( {\pi x} \right)} \right)'.\frac{{ - 1}}{{{{\cos }^2}\left( {\pi x} \right)}} = 2.\pi \frac{{\sin \left( {\pi x} \right)}}{{{{\cos }^2}\left( {\pi x} \right)}}\).

\(f'\left( 3 \right) = 2\pi .\frac{{\sin 3\pi }}{{{{\cos }^2}3\pi }} = 0\).

Hướng dẫn giải:

Chọn B.

\(y' = \left( {\cos 3x} \right)'\sin 2x + \cos 3x\left( {\sin 2x} \right)' = - 3\sin 3x.\sin 2x + 2\cos 3x.\cos 2x\).

\(y'\left( {\frac{\pi }{3}} \right) = - 3\sin 3\frac{\pi }{3}.\sin 2\frac{\pi }{3} + 2\cos 3\frac{\pi }{3}.\cos 2\frac{\pi }{3} = 1\).

Hướng dẫn giải:

Chọn D.

\(y' = \frac{{\left( {\cos 2x} \right)'.\left( {1 - \sin x} \right) - \cos 2x\left( {1 - \sin x} \right)'}}{{{{\left( {1 - \sin x} \right)}^2}}} = \frac{{ - 2\sin 2x\left( {1 - \sin x} \right) + \cos 2x.cosx}}{{{{\left( {1 - \sin x} \right)}^2}}}\).

\[y'\left( {\frac{\pi }{6}} \right) = \frac{{ - 2.\frac{{\sqrt 3 }}{2}\left( {1 - \frac{1}{2}} \right) + \frac{1}{2}.\frac{{\sqrt 3 }}{2}}}{{{{\left( {1 - \frac{1}{2}} \right)}^2}}} = \frac{{ - \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{4}}}{{\frac{1}{4}}} = 4\left( { - \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{4}} \right) = - 2\sqrt 3 + \sqrt 3 = - \sqrt 3 \].

Hướng dẫn giải:

Chọn A.

\(f'\left( x \right) = \frac{1}{{2\sqrt x }}\cos \sqrt x - \frac{1}{{2\sqrt x }}\sin \sqrt x = \frac{1}{{2\sqrt x }}\left( {\cos \sqrt x - \sin \sqrt x } \right)\).

\(f'\left( {\frac{{{\pi ^2}}}{{16}}} \right) = \frac{1}{{2\sqrt {{{\left( {\frac{\pi }{4}} \right)}^2}} }}\left( {\cos \sqrt {{{\left( {\frac{\pi }{4}} \right)}^2}} - \sin \sqrt {{{\left( {\frac{\pi }{4}} \right)}^2}} } \right) = \frac{1}{{2.\frac{{\sqrt 2 }}{2}}}\left( {\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}} \right) = 0\).

Hướng dẫn giải:

Chọn C.

\(y = \sqrt {\tan x + \cot x} \Rightarrow {y^2} = \tan x + \cot x \Rightarrow y'.2y = \frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}\).

\( \Rightarrow y' = \frac{1}{{2\sqrt {\tan x + \cot x} }}\left( {\frac{1}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right)\).

\(f'\left( {\frac{\pi }{4}} \right) = \frac{1}{{2\sqrt {\tan \frac{\pi }{4} + \cot \frac{\pi }{4}} }}\left( {\frac{1}{{{{\cos }^2}\left( {\frac{\pi }{4}} \right)}} - \frac{1}{{{{\sin }^2}\left( {\frac{\pi }{4}} \right)}}} \right) = \frac{1}{{2\sqrt 2 }}\left( {2 - 2} \right) = 0\)