Cho hàm số \(y = f\left( x \right) = - \frac{1}{x}\). Xét hai mệnh đề :

\(\left( I \right):y'' = f''\left( x \right) = \frac{2}{{{x^3}}}\).\(\left( {II} \right):y''' = f'''\left( x \right) = - \frac{6}{{{x^4}}}\).

Mệnh đề nào đúng?

A. \[y'' = - \frac{{2\sin x}}{{{{\cos }^3}x}}\].

B. \[y'' = \frac{1}{{{{\cos }^2}x}}\].

C. \[y'' = - \frac{1}{{{{\cos }^2}x}}\].

D. \[y'' = \frac{{2\sin x}}{{{{\cos }^3}x}}\].

A.$dy=4\cos 2x\sin 2xdx$

B.$dy=2\cos 2x\sin 2xdx$

C.$dy=-2\cos 2x\sin 2xdx$

D.$dy=-2\sin 4xdx$

A. \[x = \frac{\pi }{2}\].

B. \(x = 0\) và \[x = \frac{\pi }{6}\].

C. \(x = 0\) và \[x = \frac{\pi }{3}\].

D. \(x = 0\) và \[x = \frac{\pi }{2}\].

A.$dy=\frac{1}{2\sqrt{x}{\cos }^{2}x}dx$

B.$dy=\frac{1}{\sqrt{x}{\cos }^2\sqrt{x}}dx$

C. .$dy=\frac{1}{2\sqrt{x}\cos \sqrt{x}}dx$

D.$dy=\frac{1}{2\sqrt{x}{\cos }^2\sqrt{x}}dx$

A. \[0\].

B. \[ - 1\].

C. \( - 2\).

D. \(5\).

A.$df\left(x\right)=\frac{-\sin \left(4x\right)}{2\sqrt{1+{\cos }^{2}2x}}dx$

B.$df\left(x\right)=\frac{-\sin \left(4x\right)}{\sqrt{1+{\cos }^{2}2x}}dx$

C.$df\left(x\right)=\frac{\cos \left(2x\right)}{\sqrt{1+{\cos }^{2}2x}}dx$

D.$df\left(x\right)=\frac{-\sin \left(2x\right)}{\sqrt{1+{\cos }^{2}2x}}dx$

A.$d\left[f\left(x\right)\right]=\frac{\sin 2x}{2\sqrt{\cos 2x}}dx$

B.$d\left[f\left(x\right)\right]=\frac{\sin 2x}{\sqrt{\cos 2x}}dx$

C.$d\left[f\left(x\right)\right]=\frac{-\sin 2x}{2\sqrt{\cos 2x}}dx$

D.$d\left[f\left(x\right)\right]=\frac{-\sin 2x}{\sqrt{\cos 2x}}dx$