Cho hàm số y=cos2x/(1-sinx) , tính y'(pi/6) bằng

* Hướng dẫn giải

Chọn D

$\begin{array}{l}y\text{'}=\frac{\left(\cos 2x\right)\text{'}.\left(1-\sin x\right)-\cos 2x\left(1-\sin x\right)\text{'}}{{\left(1-\sin x\right)}^2}\\ =\frac{-2\sin 2x\left(1-\sin x\right)+\cos 2x.cosx}{{\left(1-\sin x\right)}^2}\end{array}$

$\begin{array}{l}y\text{'}\left(\frac{\pi }{6}\right)=\frac{-2.\frac{\sqrt{3}}{2}\left(1-\frac{1}{2}\right)+\frac{1}{2}.\frac{\sqrt{3}}{2}}{{\left(1-\frac{1}{2}\right)}^2}\\ =\frac{-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}}{\frac{1}{4}}\\ =4\left(-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}\right)\\ =-2\sqrt{3}+\sqrt{3}=-\sqrt{3}\end{array}$