Tính đạo hàm của hàm số sau y=( sin x + cos x) / (sin x - cos x) -2 sin x/ (sin x-cos x) mũ 2

* Hướng dẫn giải

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

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

$=\frac{-2}{{(\sin x-\cos x)}^2}$

Chọn đáp án C