Cho hàm số \(y=\cos3x\sin2x\). Tính \(y'\left(\dfrac{\pi}{3}\right)\) .
\(-1\).\(-\dfrac{1}{2}\).\(\dfrac{1}{2}\).\(1\).Hướng dẫn giải:\(y=\cos3x\sin2x=\dfrac{1}{2}\left(\sin\left(2x-3x\right)+\sin\left(2x+3x\right)\right)=\dfrac{1}{2}\left(\sin5x-\sin x\right)\), suy ra \(y'=\dfrac{1}{2}\left(5\cos5x-\cos x\right)\) và
\(y'\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}\left(5\cos\dfrac{5\pi}{3}-\cos\dfrac{\pi}{3}\right)=2\cos\dfrac{\pi}{3}=1.\)