Chọn D.

Đáp án là C.

Lời giải:

$y=\sqrt{\cos(x^2+2x+3)}$

$y'=\frac{-(x+1)\sin (x^2+2x+3)}{\sqrt{\cos (x^2+2x+3)}}$

Chỉ em phương pháp múa cột trong tính nguyên hàm với chị

\(y'=-2\left(\dfrac{\pi}{4}-2x\right)'sin\left(\dfrac{\pi}{4}-2x\right)\\ =4sin\left(\dfrac{\pi}{4}-2x\right)\)

xét hàm số y=\(\sin\left(cos^2x\right)cos\left(sin^2x\right)\) =

\(\frac{sin\left(cos^2x+sin^2x\right)+sin\left(cos^2x-sin^2x\right)}{2}=\frac{sin1+sin\left(cós2x\right)}{2}\)

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

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

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

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

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

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

tham khảo:

a)\(y'=xsin2x+sin^2x\)

\(y'=sin^2x+xsin2x\)

b)\(y'=-2sin2x+2cosx\\ y'=2\left(cosx-sin2x\right)\)

c)\(y=sin3x-3sinx\)

\(y'=3cos3x-3cosx\)

d)\(y'=\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\)

\(y'=\dfrac{sin^2x-cos^2x}{sin^2x.cos^2x}\)

y ' = ( sin x + cos x ) ' y ' = ( sin x ) ' + ( cos x ) ' = c osx - sinx 

 Chọn đáp án C