cho hàm số y= $\dfrac{cos2x}{1-sinx}$ => y'( $\dfrac{\pi}{6}$ )=?

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1. Đáp án: $-\sqrt3$

    

   Giải thích các bước giải:

   $y=\dfrac{\cos2x}{1-\sin x}$

   $y’=\dfrac{(\cos2x)'(1-\sin x)-\cos2x(1-\sin x)’}{(1-\sin x)^2}$

   $=\dfrac{-2\sin2x(1-\sin x)+\cos x\cos2x}{(1-\sin x)^2}$

   $=\dfrac{2\sin x\sin2x+\cos x\cos2x-2\sin2x}{(1-\sin x)^2}$

   Thay $x=\dfrac{\pi}{6}$ vào $y’$

   $\to y’\Big(\dfrac{\pi}{6}\Big)=-\sqrt3$

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2. Đáp án:

   \(f’\left(\dfrac{\pi}{6}\right)=-\sqrt3\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   \quad y = f(x) = \dfrac{\cos2x}{1-\sin x}\\
   \to y’ = f'(x) = \dfrac{(\cos2x)'(1-\sin x) – (1-\sin x)’\cos2x}{(1-\sin x)^2}\\
   \to y’ = f'(x) = \dfrac{-2\sin2x(1-\sin x) +\cos x\cos2x}{(1-\sin x)^2}\\
   \to f’\left(\dfrac{\pi}{6}\right)=\dfrac{-2\sin\dfrac{\pi}{3}\left(1-\sin\dfrac{\pi}{6}\right) +\cos\dfrac{\pi}{6}\cos\dfrac{\pi}{3}}{\left(1-\sin\dfrac{\pi}{6}\right)^2}\\
   \to f’\left(\dfrac{\pi}{6}\right)=\dfrac{-2\cdot\dfrac{\sqrt3}{2}\cdot\left(1 – \dfrac12\right) + \dfrac{\sqrt3}{2}\cdot \dfrac12}{\left(1 – \dfrac12\right)^2}\\
   \to f’\left(\dfrac{\pi}{6}\right)=-\sqrt3
   \end{array}\) 

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