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. \[0\].

B. \[ - 1\].

C. \( - 2\).

D. \(5\).

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. \[\frac{1}{{\cos x}}\].

B. \[ - \frac{1}{{\cos x}}\].

C. \(\cot x\).

D. \(tanx\).

A.$df\left(0\right)=-dx$

B.$f^{\text{'}}\left(0^{+}\right)=\underset{x\to 0^{+}}{\lim } \frac{x^2-x}{x}=\underset{x\to 0^{+}}{\lim }(x-1)=-1$

C.$f^{\text{'}}\left(0^{+}\right)=\underset{x\to 0^{+}}{\lim }(x^2-x)=0$

D.$f^{\text{'}}\left(0^{-}\right)=\underset{x\to 0^{-}}{\lim }2x=0$

A.$dy=\frac{-8}{{(2x-1)}^2}dx$

B.$dy=\frac{4}{{(2x-1)}^2}dx$

C.$dy=\frac{-4}{{(2x-1)}^2}dx$

D.$dy=\frac{-7}{{(2x-1)}^2}dx$

A. \[y''' = {\rm{ }}12\left( {{x^2} + {\rm{ }}1} \right)\].

B. \[y''' = {\rm{ }}24\left( {{x^2} + {\rm{ }}1} \right)\].

C. \[y''' = {\rm{ }}24\left( {5{x^2} + {\rm{ }}3} \right)\].

D. \[y''' = {\rm{ }}-12\left( {{x^2} + {\rm{ }}1} \right)\].

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

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

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

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