Cho hàm số \(y = \frac{{x + 2}}{{x - 1}}\). Vi phân của hàm số là:

\({\rm{d}}y = - \frac{{{x^2} - 2x - 2}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

\({\rm{d}}y = \frac{{2x + 1}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

\({\rm{d}}y = - \frac{{2x + 1}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

\({\rm{d}}y = \frac{{{x^2} - 2x - 2}}{{{{(x - 1)}^2}}}{\rm{d}}x\)

\({\rm{d}}y = \left( { - \cos x + 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = \left( { - \cos x - 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = \left( {\cos x + 3\sin x} \right){\rm{d}}x\)

\({\rm{d}}y = - \left( {\cos x + 3\sin x} \right){\rm{d}}x\)

\[{\rm{d}}y = -\sin 2x\,{\rm{d}}x\]

\[{\rm{d}}y = \sin 2x\,{\rm{d}}x\]

\[{\rm{d}}y = \sin x\,{\rm{d}}x\]

\[{\rm{d}}y = {\rm{2cos}}x\,{\rm{d}}x\]

\[{\rm{d}}y = \left( {x\cos x-\sin x} \right){\rm{d}}x\]

\[{\rm{d}}y = \left( {x\cos x} \right){\rm{d}}x\]

\[{\rm{d}}y = \left( {\cos x-\sin x} \right){\rm{d}}x\].

\[{\rm{d}}y = \left( {x\sin x} \right){\rm{d}}x\]

\[dy = \frac{{1 - {x^2}}}{{{{({x^2} + 1)}^2}}}dx\]

\[dy = \frac{{2x}}{{({x^2} + 1)}}dx\]

\[dy = \frac{{1 - {x^2}}}{{({x^2} + 1)}}dx\]

\[dy = \frac{1}{{{{({x^2} + 1)}^2}}}dx\]

.$dy=2(x-1)dx$

$dy=2(x-1)$

$dy=(x-1)dx$

$dy={(x-1)}^{2}dx$

$dy=\frac{1}{7}dx$

$dy=7dx$

$dy=\frac{-1}{7}dx$

$dy=-7dx$

9

-9

90

-90

$dy=\cos \left(sinx\right)sinxdx$

$dy=\sin \left(\cos x\right)dx$

$dy=\cos \left(sinx\right)cosxdx$

$dy=\cos \left(sinx\right)dx$

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

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

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

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