Cho hàm số \(f\left(x\right)=\dfrac{\cos x}{1-\sin x}\). Tính \(f'\left(\dfrac{\pi}{6}\right)-f'\left(-\dfrac{\pi}{6}\right)\) .
\(\dfrac{4}{3}\).\(\dfrac{4}{9}\).\(\dfrac{8}{9}\).\(\dfrac{2}{5}\).Hướng dẫn giải:Đặt \(u=\cos x,v=1-\sin x\) thì
\(u'=-\sin x,v'=-\cos x,u'v-uv'=-\sin x\left(1-\sin x\right)-\cos x.\left(-\cos x\right)=1-\sin x\)
\(f'\left(x\right)=\dfrac{u'v-uv'}{v^2}=\dfrac{1-\sin x}{\left(1-\sin x\right)^2}=\dfrac{1}{1-\sin x}\)
\(f'\left(\dfrac{\pi}{6}\right)-f'\left(-\dfrac{\pi}{6}\right)=\dfrac{1}{1-\dfrac{1}{2}}-\dfrac{1}{1+\dfrac{1}{2}}=2-\dfrac{2}{3}=\dfrac{4}{3}\)