\(a,y'=8x^3-10x\\ \Rightarrow y''=24x^2-10\\ b,y'=e^x+xe^x\\ \Rightarrow y''=e^x+e^x+xe^x=2e^x+xe^x\)

\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)

\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)

\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)

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

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

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

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

a: \(y=x\cdot e^{2x}\)

=>\(y'=\left(x\cdot e^{2x}\right)'\)

\(=x\cdot\left(e^{2x}\right)'+x'\cdot\left(e^{2x}\right)\)

\(=e^{2x}+2\cdot x\cdot e^{2x}\)

\(y''=\left(e^{2x}+2\cdot x\cdot e^{2x}\right)'\)

\(=\left(e^{2x}\right)'+\left(2\cdot x\cdot e^{2x}\right)'\)

\(=4\cdot e^{2x}+4\cdot x\cdot e^{2x}\)

b: \(y=ln\left(2x+3\right)\)

=>\(y'=\dfrac{\left(2x+3\right)'}{\left(2x+3\right)}=\dfrac{2}{2x+3}\)

=>\(y''=\left(\dfrac{2}{2x+3}\right)'=\dfrac{2\left(2x+3\right)'-2'\left(2x+3\right)}{\left(2x+3\right)^2}\)

\(=\dfrac{4}{\left(2x+3\right)^2}\)

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

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

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, \(y=\left(2x^3+3\right)^2\)

\(y'=2\left(2x^3+3\right)6x^2\)

\(=12x^2\left(2x^3+3\right)\)

b,\(y=cos3x\)

\(y'=-3sin3x\)

c, \(y=log_2\left(x^2+2\right)\)

\(y'=\dfrac{2x}{\left(x^2+2\right)ln2}\)