Toán cho hàm số y= $\dfrac{cos2x}{1-sinx}$ => y'( $\dfrac{\pi}{6}$ )=? 13/09/2021 By Lydia cho hàm số y= $\dfrac{cos2x}{1-sinx}$ => y'( $\dfrac{\pi}{6}$ )=?
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 $y’$ $\to y’\Big(\dfrac{\pi}{6}\Big)=-\sqrt3$ Trả lời
Đáp án: \(f’\left(\dfrac{\pi}{6}\right)=-\sqrt3\) Giải thích các bước giải: \(\begin{array}{l}\quad y = f(x) = \dfrac{\cos2x}{1-\sin x}\\\to y’ = f'(x) = \dfrac{(\cos2x)'(1-\sin x) – (1-\sin x)’\cos2x}{(1-\sin x)^2}\\\to y’ = f'(x) = \dfrac{-2\sin2x(1-\sin x) +\cos x\cos2x}{(1-\sin x)^2}\\\to f’\left(\dfrac{\pi}{6}\right)=\dfrac{-2\sin\dfrac{\pi}{3}\left(1-\sin\dfrac{\pi}{6}\right) +\cos\dfrac{\pi}{6}\cos\dfrac{\pi}{3}}{\left(1-\sin\dfrac{\pi}{6}\right)^2}\\\to f’\left(\dfrac{\pi}{6}\right)=\dfrac{-2\cdot\dfrac{\sqrt3}{2}\cdot\left(1 – \dfrac12\right) + \dfrac{\sqrt3}{2}\cdot \dfrac12}{\left(1 – \dfrac12\right)^2}\\\to f’\left(\dfrac{\pi}{6}\right)=-\sqrt3\end{array}\) Trả lời
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 $y’$
$\to y’\Big(\dfrac{\pi}{6}\Big)=-\sqrt3$
Đáp án:
\(f’\left(\dfrac{\pi}{6}\right)=-\sqrt3\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad y = f(x) = \dfrac{\cos2x}{1-\sin x}\\
\to y’ = f'(x) = \dfrac{(\cos2x)'(1-\sin x) – (1-\sin x)’\cos2x}{(1-\sin x)^2}\\
\to y’ = f'(x) = \dfrac{-2\sin2x(1-\sin x) +\cos x\cos2x}{(1-\sin x)^2}\\
\to f’\left(\dfrac{\pi}{6}\right)=\dfrac{-2\sin\dfrac{\pi}{3}\left(1-\sin\dfrac{\pi}{6}\right) +\cos\dfrac{\pi}{6}\cos\dfrac{\pi}{3}}{\left(1-\sin\dfrac{\pi}{6}\right)^2}\\
\to f’\left(\dfrac{\pi}{6}\right)=\dfrac{-2\cdot\dfrac{\sqrt3}{2}\cdot\left(1 – \dfrac12\right) + \dfrac{\sqrt3}{2}\cdot \dfrac12}{\left(1 – \dfrac12\right)^2}\\
\to f’\left(\dfrac{\pi}{6}\right)=-\sqrt3
\end{array}\)