12 câu Trắc nghiệm Đại số và Giải tích 11 Bài 4 (Có đáp án): Vi phân

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

$dy=\frac{2x\tan \sqrt{x^2+1}}{\sqrt{x^2+1} {\sin }^2\sqrt{x^2+1}}dx$

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

$dy=\frac{2\tan \sqrt{x^2+1}}{\sqrt{x^2+1} {\sin }^2\sqrt{x^2+1}}dx$

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

$dy= 3{\cos }^2(1-x).\sin (1-x)dx$

$dy=- 3{\cos }^2(1-x)\sin (1-x)dx$

$dy=3 {\cos }^2(1-x)dx$

∆f(1)=0,11;df(1)=0,2

∆f(1)=0,11;df(1)=0,1

∆f(1)=0,2;df(1)=0,11

∆f(1)=0,2;df(1)=0,1

$dy=10{(2x+1)}^4$

$dy=5{(2x+1)}^4 dx$

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

$dy=10{(2x+1)}^4 dx$

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

$dy= 3{\cos }^2(1-x).\sin (1-x)dx$

$dy=- 3{\cos }^2(1-x)\sin (1-x)dx$

$dy=3 {\cos }^2(1-x)dx$

$\text{d}y=\frac{2\sqrt{x}}{4x\sqrt{x}{\cos }^2\sqrt{x}}\text{d}x$

$\text{d}y=\frac{\sin \left(2\sqrt{x}\right)}{4x\sqrt{x}{\cos }^2\sqrt{x}}\text{d}x$

$\text{d}y=\frac{2\sqrt{x}-\sin \left(2\sqrt{x}\right)}{4x\sqrt{x}{\cos }^2\sqrt{x}}\text{d}x$

$\text{d}y=-\frac{2\sqrt{x}-\sin \left(2\sqrt{x}\right)}{4x\sqrt{x}{\cos }^2\sqrt{x}}\text{d}x$

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

$dy=7dx$

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

$dy=-7dx$

9

-9

90

-90

-0,02

0,01

0,4

-0,06

$\frac{\sqrt{2}}{2}+\frac{\pi }{180}$

$\frac{1}{2}+\frac{\sqrt{2}\pi }{240}$

$\frac{\sqrt{2}}{2}+\frac{\sqrt{2}\pi }{360}$

Đáp án khác

1,0004

1,0035

1,00037

1,0005