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

Question

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

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3 tháng 2021-09-13T09:09:16+00:00 2 Answers 4 views 0

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 $y’$

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

2. Đáp án:

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

Giải thích các bước giải:

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