Vi phân của $y=\tan \left(5x\right)$ là :

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

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

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

$dy=-2\sin 4xdx$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[x = \frac{\pi }{2}\]

\(x = 0\) và \[x = \frac{\pi }{6}\]

\(x = 0\) và \[x = \frac{\pi }{3}\]

\(x = 0\) và \[x = \frac{\pi }{2}\]

Chỉ \[\left( I \right)\] đúng

Chỉ \[\left( {II} \right)\] đúng

Cả hai đều đúng

Cả hai đều sai

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

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

\(\cot x\)

\(tanx\)