Trắc nghiệm Các quy tắc tính đạo hàm có đáp án (phần 2) !!

A. 2 

B. 6 

C. - 4 

D. 3 

A. 4

B. 14

C. 15

D. 24

A. -32 

B.30 

C. - 64 

D. 12 

A.{-1;2}

B.{-1;3}

C. {0;4}

D.{1;2}

A. 1

B. -3

C. -5

D. 0 

A. 0 

B. 2 

C. 1 

D. Không tồn tại

A.$\left[-\sqrt{3};\sqrt{3}\right]$

B.$\left[-\frac{1}{\sqrt{3}};\frac{1}{\sqrt{3}}\right]$

C.$(-\infty ;-\sqrt{3}] \cup [ \sqrt{3};+\infty )$

D.$(-\infty ;-\frac{1}{\sqrt{3}}]\cup [\frac{1}{\sqrt{3}};+\infty )$

A. $m\le \sqrt{2}$

B.$m\le 2$

C.$m\le 0$

D. m<0

A. y'(1)= -4

B.y'(1)= -3

C.y'(1)= -2

D.y'(1)= -5

A.$y^{\text{'}}=2016{{(x^3-2x^2)}^2}^{015}.$

B.$y^{\text{'}}=2016{(x^3-2x^2)}^{2015}(3x^2-4x).$

C.$y^{\text{'}}=2016(x^3-2x^2)(3x^2-4x).$

D.$y^{\text{'}}=2016(x^3-2x^2)(3x^2-2x).$

A.$\frac{-3}{{\left(x+2\right)}^2}$

B.$\frac{3}{\left(x+2\right)}$

C.$\frac{3}{{\left(x+2\right)}^2}$

D.$\frac{2}{{\left(x+2\right)}^2}$

A.$-1-\frac{3}{{\left(x-2\right)}^2}$

B.$1+\frac{3}{{\left(x-2\right)}^2}$

C.$-1+\frac{3}{{\left(x-2\right)}^2}$

D.$1-\frac{3}{{\left(x-2\right)}^2}$

A.$\frac{-3x^2-13x-10}{{\left(x^2+3\right)}^2}$

B.$\frac{-x^2+x+3}{{\left(x^2+3\right)}^2}$

C.$\frac{-x^2+2x+3}{{\left(x^2+3\right)}^2}$

D.$\frac{-7x^2-13x-10}{{\left(x^2+3\right)}^2}$

A.$\frac{3x-1}{\sqrt{3x^2-2x+1}}$

B.$\frac{6x-2}{\sqrt{3x^2-2x+1}}$

C.$\frac{3x^2-1}{\sqrt{3x^2-2x+1}}$

D.$\frac{1}{2\sqrt{3x^2-2x+1}}$

A.$y\text{'}=\frac{2x-2}{\sqrt{x^2-2x}}$

B.$y\text{'}=\frac{3x^2-4}{\sqrt{x^2-2x}}$

C.$y\text{'}=\frac{2x^2-3x}{\sqrt{x^2-2x}}$

D.$y\text{'}=\frac{2x^2-2x-1}{\sqrt{x^2-2x}}$

A.$x\in \left(-\infty ;1\right)\cup \left(3;+\text{}\infty \right)$

B.$x\in \left(-\infty ;0\right)\cup \left(1;+\text{}\infty \right)$

C.$x\in \left(0;3\right)$

D.$x\in \left(0;1\right)\cup \left(1;3\right)$

A.$\frac{-x}{\left(x^2+2\right)\sqrt{x^2+2}}$

B.$\frac{x+8}{\left(x^2+2\right)\sqrt{x^2+2}}$

C.$\frac{-x-8}{\left(x^2+2\right)\sqrt{x^2+2}}$

D.$\frac{-x+8}{\left(x^2+2\right)\sqrt{x^2+2}}$

A.$\text{}S=\left\{0;1\right\}$

B.$\text{}S=\left\{0;-1\right\}$

C.$\text{}S=\left\{1;-1\right\}$

D.$\text{}S=\left\{-1;0;1\right\}$

A.$-\frac{9{\left(2x+1\right)}^2}{{\left(x-1\right)}^4}$

B.$\frac{3{\left(2x+1\right)}^2}{{\left(x-1\right)}^5}$

C.$\frac{3{\left(3x+1\right)}^2}{{\left(x-1\right)}^4}$

D.$\frac{2{\left(x+1\right)}^2}{{\left(x-1\right)}^6}$

A.$-\frac{1}{{\left(x^2-x+1\right)}^{10}}$

B.$\frac{5\left(2x-1\right)}{{\left(x^2-x+1\right)}^{10}}$

C.$-\frac{5\left(2x-1\right)}{{\left(x^2-x+1\right)}^6}$

D.$-\frac{5\left(2x-1\right)}{{\left(x^2-x+1\right)}^{10}}$

A.$\frac{2-2x}{\sqrt{1+2x-x^2}}$

B.$\frac{1-x}{\sqrt{1+2x-x^2}}$

C.$\frac{1-x}{2\sqrt{1+2x-x^2}}$

D.$\frac{1}{2\sqrt{1+2x-x^2}}$

A.$\frac{1}{2\sqrt{\frac{x^2+1}{x}}}\left(1-\frac{1}{x^2}\right)$

B.$\frac{1}{\sqrt{\frac{x^2+1}{x}}}\left(1+\frac{1}{x^2}\right)$

C.$\frac{2}{\sqrt{\frac{x^2+1}{x}}}\left(x+\frac{1}{x^2}\right)$

D.Đáp án khác 

A.$\frac{3-x}{\sqrt{1-x}\left(1-x\right)}.$

B.$\frac{2(3-x)}{\sqrt{1-x}\left(1-x\right)}.$

C.$\frac{3-x}{2\sqrt{1-x}\left(1-x\right)}.$

D.$\frac{3-x}{2\left(1-x\right)}.$

A.$\frac{3}{2}\sqrt{x-2}$

B.$\frac{3}{4}\sqrt{x-2}.(x-2)$

C.$\frac{3}{4}(x-2)$

D.$\frac{3}{4}\sqrt{x-2}.{(x-2)}^2$

A.$x\in \left(0;1\right)$

B.$x\in \left(-1;0\right)$

C.$x\in \left(-\infty ;0\right)\cup \left(1;+\infty \right)$

D.$x\in \left(-\infty ;-1\right)\cup \left(0;+\infty \right)$

A. - 4

B. - 5 

C. - 3 

D. -2 

A.$\frac{1}{2}$

B.$\frac{1}{3}$

C.1

D.2.

A.$x=\frac{1}{8}$

B.$x=\frac{1}{2}$

C.$x=\frac{1}{64}$

D.$x=\frac{1}{4}$