Cho hàm số \[y = {\sin ^2}x\]. Vi phân của hàm số là:

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

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

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

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

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

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

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

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

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

$dy=2(x-1)$

$dy=(x-1)dx$

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

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

$dy=7dx$

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

$dy=-7dx$

9

-9

90

-90

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

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

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

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

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

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

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

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

$f^{\text{'}}\left(0^{+}\right)=1$

$f^{\text{'}}\left(0^{-}\right)=1$

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

Hàm số không có vi phân tại x=0

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

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

$dy=\frac{-4}{1+x^2}dx$

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

\(y'' = 0\)

\(y'' = \frac{1}{{{{\left( {x - 2} \right)}^2}}}\)

\(y'' = - \frac{4}{{{{\left( {x - 2} \right)}^2}}}\)

\(y'' = \frac{4}{{{{\left( {x - 2} \right)}^3}}}\)