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