Cho hàm số \(y = f\left( x \right) = \frac{1}{{\sqrt {\sin x} }}\). Giá trị \(f'\left( {\frac{\pi }{2}} \right)\) bằng:

Hướng dẫn giải:

Chọn C.

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

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

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