Trắc nghiệm Toán 10 Bài 5 Đạo hàm cấp hai

A. 9x2-12x+sinx

B. 9x2-12x-sinx

C. 9x2-6x-sinx

D. 9x2-12x+cosx

A. \(\frac{8}{{{{\left( {2x - 3} \right)}^3}}}\)

B. \(\frac{{ - 8}}{{{{\left( {2x - 3} \right)}^2}}}\)

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

D. \(\frac{{ - 4}}{{{{\left( {2x - 3} \right)}^2}}}\)

A. cos2x+ 4cos4x +9cos6x

B. –cos2x -4cos4x – 9cos6x

C. –cosx-4cos2x-9cos3x

D. \( - \frac{1}{4}{\rm{cos}}2x + \frac{1}{4}{\rm{cos}}4x - \frac{1}{4}{\rm{cos}}6x\)

A. 10

B. 8

C. 18

D. 1

A. -4cos2x

B. 4cos2x

C. -2sin2x

D. -4sin2x

A. y''+y=-2cosx

B. y''-y'=2cosx

C. y''+y'=2cosx

D. y''+y=2cosx

A. \(16{x^3} - 6x\)

B. \(4{x^3} - 6\)

C. \(16{x^3} - 6\)

D. \(16{x^2} - 6\)

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

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

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

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

A. \(\frac{{2\tan x}}{{{{\cos }^2}x}} - \frac{{2\cot x}}{{{{\sin }^2}x}} - \sin x + \cos x\)

B. 0

C. \({\tan ^2}x - {\cot ^2}x + \cos x - \sin x\)

D. \(\frac{{2\tan x}}{{{{\cos }^2}x}} + \frac{{2\cot x}}{{{{\sin }^2}x}} - \sin x - \cos x\)

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

B. \(4y + y'' = 0\)

C. \(4y - y'' = 0\)

D. \(y = y'\tan 2x\)

A. \(\frac{{2\left( {7{x^3} + 15{x^2} - 93x + 77} \right)}}{{{{\left( {{x^2} - 2x - 3} \right)}^3}}}\)

B. \(\frac{{2\left( {7{x^3} - 15{x^2} + 93x - 77} \right)}}{{{{\left( {{x^2} - 2x - 3} \right)}^3}}}\)

C. \(\frac{{2\left( {7{x^3} + 15{x^2} + 93x - 77} \right)}}{{{{\left( {{x^2} - 2x - 3} \right)}^3}}}\)

D. \(\frac{{2\left( {7{x^3} - 15{x^2} - 93x + 77} \right)}}{{{{\left( {{x^2} - 2x - 3} \right)}^3}}}\)

A. 0

B. 2

C. 4cosx

D. 6sinx+4cosx

A. y=3x2

B. y=2x3

C. y=x3

D. y=x2

A. -2

B. \(2\sqrt 3 \)

C. -4

D. \(-2\sqrt 3 \)

A. Chỉ (I)

B. Chỉ (II)

C. Cả 2 đều sai

D. Cả 2 đều đúng