Hàm số \(y{\rm{ }} = \frac{x}{{{x^2} + 1}}\). Có vi phân là:

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

$dy=2(x-1)$

$dy=(x-1)dx$

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

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

$dy=7dx$

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

$dy=-7dx$

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

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

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

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

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

\(y'' = 0\)

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

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

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

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

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

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

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

\(y'' = \frac{1}{{(2x + 5)\sqrt {2x + 5} }}\)

\(y'' = \frac{1}{{\sqrt {2x + 5} }}\)

\(y'' = - \frac{1}{{(2x + 5)\sqrt {2x + 5} }}\)

\(y'' = - \frac{1}{{\sqrt {2x + 5} }}\)