Tính đạo hàm của hàm số sau y=(sin2x+cos2x)/(2sin2x-cos2x)

* Hướng dẫn giải

Chọn D

$y\text{'}=\frac{{\left(\sin 2x+\cos 2x\right)}^{/}.\left(2\sin 2x-\cos 2x\right)-{\left(2\sin 2x-\cos 2x\right)}^{/}.\left(\sin 2x+\cos 2x\right)}{{\left(2\sin 2x-\cos 2x\right)}^2}$

$\begin{array}{l}y\text{'}=\frac{\left(2\cos 2x-2\sin 2x\right)\left(2\sin 2x-\cos 2x\right)-\left(4\cos 2x+2\sin 2x\right)\left(\sin 2x+\cos 2x\right)}{{\left(2\sin 2x-\cos 2x\right)}^2}\\ =\frac{4.c{\text{os2x. sin2x - 2cos}}^{2}2x-4{\sin }^{2}2x+\text{}2.\sin 2x.c\text{os2x }}{{(2\sin 2x-\cos 2x)}^2}\\ -\frac{{\text{ ( 4cos2x . sin2x + 4cos}}^{2}2x+2{\sin }^{2}2x+2\sin 2x.c\text{os2x}}{{(\text{}2\sin 2x-c\text{os2x)}}^2}\end{array}$

$\begin{array}{l}y\text{'}=\frac{-6{\cos }^{2}2x-6{\sin }^{2}2x}{{\left(2\sin 2x-\cos 2x\right)}^2}\\ =\frac{-6}{{\left(2\sin 2x-\cos 2x\right)}^2}\end{array}$