Đạo hàm

Câu hỏi: Đạo hàm của hàm số $y=\sqrt{2+\tan \left(x+\frac{1}{x}\right)}$ là:

A.$y^{\text{'}}=\frac{1}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}\cdot$

B.$y^{\text{'}}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}\cdot$

C.$y^{\text{'}}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}.\left(1-\frac{1}{x^2}\right).$

D.$y^{\text{'}}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}.\left(1+\frac{1}{x^2}\right).$