Hàm số \(y = \frac{{ - 2{x^2} + 3x}}{{1 - x}}\) có đạo hàm cấp \(2\) bằng :

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

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

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

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

C. \(\cot x\).

D. \(tanx\).

A. \[y' = \sin \left( {x + \frac{\pi }{2}} \right)\].

B. \[y'' = \sin \left( {x + \pi } \right)\].

C. \[y''' = \sin \left( {x + \frac{{3\pi }}{2}} \right)\].

D. \[{y^{\left( 4 \right)}} = \sin \left( {2\pi - x} \right)\].

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