100+ câu trắc nghiệm Giải tích 3 có đáp án - Phần 2

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

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

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

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

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

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

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

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

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

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

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

$dy=-2\sin 4xdx$

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

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

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

Hàm số không có vi phân tại x=0.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(y'' = 0\).

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

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

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

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

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

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

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

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

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

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

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

\({y^{(5)}} = - \frac{{120}}{{{{(x + 1)}^6}}}\).

\({y^{(5)}} = \frac{{120}}{{{{(x + 1)}^6}}}\).

\({y^{(5)}} = \frac{1}{{{{(x + 1)}^6}}}\).

\({y^{(5)}} = - \frac{1}{{{{(x + 1)}^6}}}\).

\[{y^{\left( 5 \right)}} = - \frac{{120}}{{{{\left( {x + 1} \right)}^6}}}\].

\[{y^{\left( 5 \right)}} = \frac{{120}}{{{{\left( {x + 1} \right)}^5}}}\].

\[{y^{\left( 5 \right)}} = \frac{1}{{{{\left( {x + 1} \right)}^5}}}\].

\[{y^{\left( 5 \right)}} = - \frac{1}{{{{\left( {x + 1} \right)}^5}}}\].

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

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

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

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

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

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

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

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

\[y'' = - \frac{{2\sin x}}{{{{\cos }^3}x}}\].

\[y'' = \frac{1}{{{{\cos }^2}x}}\].

\[y'' = - \frac{1}{{{{\cos }^2}x}}\].

\[y'' = \frac{{2\sin x}}{{{{\cos }^3}x}}\].

\[y' = \sin \left( {x + \frac{\pi }{2}} \right)\].

\[y'' = \sin \left( {x + \pi } \right)\].

\[y''' = \sin \left( {x + \frac{{3\pi }}{2}} \right)\].

\[{y^{\left( 4 \right)}} = \sin \left( {2\pi - x} \right)\].

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

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

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

\[y'' = \frac{2}{{{{\left( {1 - x} \right)}^4}}}\].

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

\(x = 0\) và \[x = \frac{\pi }{6}\].

\(x = 0\) và \[x = \frac{\pi }{3}\].

\(x = 0\) và \[x = \frac{\pi }{2}\].

\[4y - y' = 0\].

\[4y + y'' = 0\].

\(y = y'tan2x\).

\({y^2} = {\left( {y'} \right)^2} = 4\).

Chỉ \[\left( I \right)\] đúng.

Chỉ \[\left( {II} \right)\] đúng.

Cả hai đều đúng.

Cả hai đều sai.

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

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

\(\cot x\).

\(tanx\).

Chỉ \[\left( I \right)\] đúng.

Chỉ \[\left( {II} \right)\] đúng.

Cả hai đều đúng.

Cả hai đều sai.

\[3\].

\[6\].

\(12\).

\(24\).

\[0\].

\[ - 1\].

\( - 2\).

\(5\).

\[\left[ { - 1;2} \right]\].

\[\left( { - \infty ;0} \right]\].

\(\left\{ { - 1} \right\}\).

\(\emptyset \).

\[y'''\left( 1 \right) = \frac{3}{8}\].

\[y'''\left( 1 \right) = \frac{1}{8}\].

\[y'''\left( 1 \right) = - \frac{3}{8}\].

\[y'''\left( 1 \right) = - \frac{1}{4}\].

\[{y^{\left( {10} \right)}}\left( 1 \right) = 0\].

\[{y^{\left( {10} \right)}}\left( 1 \right) = 10a + b\].

\[{y^{\left( {10} \right)}}\left( 1 \right) = 5a\].

\[{y^{\left( {10} \right)}}\left( 1 \right) = 10a\].