Vi phân của hàm số $f\left(x\right)=3x^2-x$ tại điểm x=2 , ứng với $△x=0,1$ là:

$dy=-2017\sin \left(2017x\right)dx$

$dy=\frac{2017}{{\sin }^2\left(2017x\right)}dx$

$dy=\frac{-2017}{co{s}^2\left(2017x\right)}dx$

$dy=\frac{-2017}{{\sin }^2\left(2017x\right)}dx$

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

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

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

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

$dy=\frac{5x}{{\cos }^{2}5x}$$dx$

$dy=\frac{-5x}{{\sin }^{2}5x}$$dx$

$dy=\frac{5}{co{s}^{2}5x}$$dx$

$dy=4\cos 2x\sin 2xdx$

$dy=2\cos 2x\sin 2xdx$

$dy=-2\cos 2x\sin 2xdx$

$dy=-2\sin 4xdx$

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

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

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

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

$dy=\frac{1}{2\sqrt{x}{\cos }^{2}x}dx$

$dy=\frac{1}{\sqrt{x}{\cos }^2\sqrt{x}}dx$

.$dy=\frac{1}{2\sqrt{x}\cos \sqrt{x}}dx$

$dy=\frac{1}{2\sqrt{x}{\cos }^2\sqrt{x}}dx$

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

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

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

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

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

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

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

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

\[y'' = - \frac{{2{x^3} + 3x}}{{\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }}\]

\[y'' = \frac{{2{x^2} + 1}}{{\sqrt {1 + {x^2}} }}\]

\[y'' = \frac{{2{x^3} + 3x}}{{\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }}\]

\[y'' = - \frac{{2{x^2} + 1}}{{\sqrt {1 + {x^2}} }}\]

\[y''' = 80{\left( {2x + 5} \right)^3}\]

\[y''' = 480{\left( {2x + 5} \right)^2}\]

\[y''' = - 480{\left( {2x + 5} \right)^2}\]

\[y''' = - 80{\left( {2x + 5} \right)^3}\]