Cho hàm số \(y = \frac{{\cos 2x}}\). Tính \(y'\left( {\frac{\pi }{6}} \right)\) bằng:

Hướng dẫn giải:

Chọn D.

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

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