Hàm số y=cosx/2sin^2x có đạo hàm bằng

* Hướng dẫn giải

Chọn  A

$\begin{array}{l}{y}^{\text{'}}={\left(\frac{\cos x}{2{\sin }^{2}x}\right)}^{\text{'}}\\ =\frac{2{\sin }^{2}x{\left(\cos x\right)}^{\text{'}}-\left(2{\sin }^{2}x\right)\text{'}.\cos x}{4{\sin }^{4}x}\\ =\frac{-2{\sin }^{3}x-2.2\sin x.\cos x.\cos x}{4{\sin }^{4}x}\end{array}$

$\begin{array}{l}=-\frac{{\sin }^{2}x+2{\text{cos}}^{2}x}{2{\sin }^{3}x}\\ =-\frac{({\sin }^{2}x+\text{}c{\text{os}}^{2}x)+{\text{cos}}^{2}x}{2{\sin }^{3}x}\\ =-\frac{1+{\text{cos}}^{2}x}{2{\sin }^{3}x}\end{array}$