Cho \(f\left( t \right) = \frac{{\cos t}}{{1 - \sin t}}\). Tính \(f'\left( {\frac{\pi }{6}} \right)\)
A. -2
B. -3
C. 2
D. 5
Ta có:
f\prime \left( t \right) = \frac{{\left( {\cos t} \right)\prime \left( {1 - \sin t} \right) - \cos t\left( {1 - \sin t} \right)\prime }}{{{{\left( {1 - \sin t} \right)}^2}}}\\
= \frac{{ - \sin t\left( {1 - \sin t} \right) - \cos t\left( { - \cos t} \right)}}{{{{\left( {1 - \sin t} \right)}^2}}} = \frac{{ - \sin t + {{\sin }^2}t + {{\cos }^2}t}}{{{{\left( {1 - \sin t} \right)}^2}}}\\
= \frac{{1 - \sin t}}{{{{\left( {1 - \sin t} \right)}^2}}} = \frac{1}{{1 - \sin t}}
\end{array}\)
Vậy \(f'\left( {\frac{\pi }{6}} \right) = \frac{1}{{1 - \sin \frac{\pi }{6}}} = \frac{1}{{1 - \frac{1}{2}}} = 2\)
Chọn C.
-- Mod Toán 11
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