Cho hàm số \(f\left(x\right)=\sin^3x+\cos^3x\) . Tính \(f'\left(\dfrac{\pi}{6}\right)\).
\(\dfrac{3\sqrt{3}-3}{8}\).\(\dfrac{3\sqrt{3}-6}{8}\).\(\dfrac{3\sqrt{3}-9}{8}\).\(\dfrac{3\sqrt{3}-12}{8}\).Hướng dẫn giải:
\(f\left(x\right)=\sin^3x+\cos^3x,f'\left(x\right)=3\sin^2x.\left(\sin x\right)'+3\cos^2x.\left(\cos x\right)'=3\sin^2x\cos x-3\cos^2x\sin x\)
\(=\dfrac{3}{2}\sin2x.\left(\sin x-\cos x\right)\)
\(f'\left(\dfrac{\pi}{6}\right)=\dfrac{3}{2}\sin\dfrac{\pi}{3}\left(\sin\dfrac{\pi}{6}-\cos\dfrac{\pi}{6}\right)=\dfrac{3\sqrt{3}}{4}\left(\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\right)=\dfrac{3\sqrt{3}\left(1-\sqrt{3}\right)}{8}=\dfrac{3\sqrt{3}-9}{8}\)
