Đặt u = π/2 - x thì u' = -1

Do cos⁡(π/2-x) = sin⁡x

Chọn B.

Ta có: \(y' = {\left( {\tan x} \right)^\prime } = \frac{1}{{{{\cos }^2}x}}\)

Vậy \(y'\left( {\frac{{3\pi }}{4}} \right) = \frac{1}{{{{\cos }^2}\left( {\frac{{3\pi }}{4}} \right)}} = 2\).

a) Đặt \(u = 3{\rm{x}}\) thì \(y = \sin u\). Ta có: \(u{'_x} = {\left( {3{\rm{x}}} \right)^\prime } = 3\) và \(y{'_u} = {\left( {\sin u} \right)^\prime } = \cos u\).

Suy ra \(y{'_x} = y{'_u}.u{'_x} = \cos u.3 = 3\cos 3{\rm{x}}\).

Vậy \(y' = 3\cos 3{\rm{x}}\).

b) Đặt \(u = \cos 2{\rm{x}}\) thì \(y = {u^3}\). Ta có: \(u{'_x} = {\left( {\cos 2{\rm{x}}} \right)^\prime } =  - 2\sin 2{\rm{x}}\) và \(y{'_u} = {\left( {{u^3}} \right)^\prime } = 3{u^2}\).

Suy ra \(y{'_x} = y{'_u}.u{'_x} = 3{u^2}.\left( { - 2\sin 2{\rm{x}}} \right) = 3{\left( {\cos 2{\rm{x}}} \right)^2}.\left( { - 2\sin 2{\rm{x}}} \right) =  - 6\sin 2{\rm{x}}{\cos ^2}2{\rm{x}}\).

Vậy \(y' =  - 6\sin 2{\rm{x}}{\cos ^2}2{\rm{x}}\).

c) Đặt \(u = \tan {\rm{x}}\) thì \(y = {u^2}\). Ta có: \(u{'_x} = {\left( {\tan {\rm{x}}} \right)^\prime } = \frac{1}{{{{\cos }^2}x}}\) và \(y{'_u} = {\left( {{u^2}} \right)^\prime } = 2u\).

Suy ra \(y{'_x} = y{'_u}.u{'_x} = 2u.\frac{1}{{{{\cos }^2}x}} = 2\tan x\left( {{{\tan }^2}x + 1} \right)\).

Vậy \(y' = 2\tan x\left( {{{\tan }^2}x + 1} \right)\).

d) Đặt \(u = 4 - {x^2}\) thì \(y = \cot u\). Ta có: \(u{'_x} = {\left( {4 - {x^2}} \right)^\prime } =  - 2{\rm{x}}\) và \(y{'_u} = {\left( {\cot u} \right)^\prime } =  - \frac{1}{{{{\sin }^2}u}}\).

Suy ra \(y{'_x} = y{'_u}.u{'_x} =  - \frac{1}{{{{\sin }^2}u}}.\left( { - 2{\rm{x}}} \right) = \frac{{2{\rm{x}}}}{{{{\sin }^2}\left( {4 - {x^2}} \right)}}\).

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

a: \(\left(\frac{3x^2-18x-2}{1-2x}\right)^{\prime}=\frac{\left(3x^2-18x-2\right)^{\prime}\cdot\left(1-2x\right)-\left(3x^2-18x-2\right)\left(1-2x\right)^{\prime}}{\left(1-2x\right)^2}\)

\(=\frac{\left(6x-18\right)\left(1-2x\right)+2\left(3x^2-18x-2\right)}{\left(2x-1\right)^2}=\frac{6x-12x^2-18+36x+6x^2-36x-4}{\left(2x-1\right)^2}\)

\(=\frac{-6x^2+6x-22}{\left(2x-1\right)^2}\)

\(\left(\frac{2x-3}{x+4}\right)^{\prime}=\frac{\left(2x-3\right)^{\prime}\left(x+4\right)-\left(2x-3\right)\left(x+4\right)^{\prime}}{\left(x+4\right)^2}\)

\(=\frac{2\left(x+4\right)-\left(2x-3\right)}{\left(x+4\right)^2}=\frac{2x+8-2x+3}{\left(x+4\right)^2}=\frac{11}{\left(x+4\right)^2}\)

=>y'=\(\frac{-6x^2+6x-22}{\left(2x-1\right)^2}-\frac{11}{\left(x+4\right)^2}\)

\(=\frac{\left(-6x^2+6x-22\right)\left(x^2+8x+16\right)-11\left(2x-1\right)^2}{\left(2x-1\right)^2\cdot\left(x+4\right)^2}\)

\(=\frac{-6x^4-48x^3-96x^2+6x^3+48x^2+96x-22x^2-176x-352-11\left(4x^2-4x+1\right)}{\left(2x-1\right)^2\cdot\left(x+4\right)^2}\)

\(=\frac{-6x^4-42x^3-70x^2-80x-352-44x^2+44x-11}{\left(2x-1\right)^2\cdot\left(x+4\right)^2}\)

\(=\frac{-6x^4-42x^3-114x^2-36x-363}{\left(2x-1\right)^2\cdot\left(x+4\right)^2}\)

Kết hợp với điều kiện xác định suy ra 

 (k ∈ Z)

Vậy với  (k ∈ Z)

thì 

a: y=ln(x+1)

=>\(y'=\dfrac{1}{x+1}\)

=>\(y''=\dfrac{1'\left(x+1\right)-1\left(x+1\right)'}{\left(x+1\right)^2}=\dfrac{-1}{\left(x+1\right)^2}\)

b: y=tan 2x

=>\(y'=\dfrac{2}{cos^22x}\)

=>\(y''=\left(\dfrac{2}{cos^22x}\right)'=\dfrac{-2\cdot cos^22x'}{cos^42x}=\dfrac{-2\cdot2\cdot cos2x\left(cos2x\right)'}{cos^42x}\)

\(=\dfrac{4\cdot2\cdot sin2x}{cos^32x}=\dfrac{8\cdot sin2x}{cos^32x}\)

\(y'=2\left(tan^2x\right)'+3\left[cot\left(\dfrac{\pi}{3}-2x\right)\right]'\\ =2\cdot2tanx\cdot\left(tanx\right)'+3\cdot\dfrac{-\left(\dfrac{\pi}{3}-2x\right)'}{sin^2\left(\dfrac{\pi}{3}-2x\right)}\\ =\dfrac{4tanx}{cos^2x}+\dfrac{6}{sin^2\left(\dfrac{\pi}{3}-2x\right)}\)

Đáp án C.