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

a: y=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

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

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

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

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

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

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

e,

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

f,

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

a: \(y'=\left(x^2-x\right)'=2x-1\)

\(y''=\left(2x-1\right)'=2\)

b: \(y'=\left(cosx\right)'=-sinx\)

\(y''=\left(-sinx\right)'=-cosx\)

a. \(y'=\dfrac{-1}{\left(x-1\right)}\)

b. \(y'=\dfrac{5}{\left(1-3x\right)^2}\)

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

d. \(y'=\dfrac{4x\left(x^2-2x-3\right)-2x^2\left(2x-2\right)}{\left(x^2-2x-3\right)^2}=\dfrac{-4x^2-12x}{\left(x^2-2x-3\right)^2}\)

e. \(y'=1+\dfrac{2}{\left(x-1\right)^2}=\dfrac{x^2-2x+3}{\left(x-1\right)^2}\)

g. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)

2.

a. \(y'=4\left(x^2+x+1\right)^3.\left(x^2+x+1\right)'=4\left(x^2+x+1\right)^3\left(2x+1\right)\)

b. \(y'=5\left(1-2x^2\right)^4.\left(1-2x^2\right)'=-20x\left(1-2x^2\right)^4\)

c. \(y'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{2x+1}{x-1}\right)'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{-3}{\left(x-1\right)^2}\right)=\dfrac{-9\left(2x+1\right)^2}{\left(x-1\right)^4}\)

d. \(y'=\dfrac{2\left(x+1\right)\left(x-1\right)^3-3\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)^6}=\dfrac{-x^2-6x-5}{\left(x-1\right)^4}\)

e. \(y'=-\dfrac{\left[\left(x^2-2x+5\right)^2\right]'}{\left(x^2-2x+5\right)^4}=-\dfrac{2\left(x^2-2x+5\right)\left(2x-2\right)}{\left(x^2-2x+5\right)^4}=-\dfrac{4\left(x-1\right)}{\left(x^2-2x+5\right)^3}\)

f. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)

\(a,y'=\left(\dfrac{1}{2x+3}\right)'=-\dfrac{2}{\left(2x+3\right)^2}\\ \Rightarrow y''=\dfrac{2\cdot\left[\left(2x+3\right)^2\right]'}{\left(2x+3\right)^4}=\dfrac{8}{\left(2x+3\right)^3}\\ b,y'=\left(log_3x\right)'=\dfrac{1}{xln3}\\ \Rightarrow y''=-\dfrac{1}{x^2ln3}\\ c,y'=\left(2^x\right)'=2^x\cdot ln2\\ \Rightarrow y''=2^x\cdot\left(ln2\right)^2\)

a,

\(y' = 6x - 4 \Rightarrow y'' = 6\)

Tại \({x_0} =  - 2 \Rightarrow y''( - 2) = 6\)

b,

\(\begin{array}{l}y' = \frac{2}{{\left( {2x + 1} \right)\ln 3}}\\ \Rightarrow y'' = \left( {2.\frac{1}{{\left( {\left( {2x + 1} \right)\ln 3} \right)}}} \right)' =  - 2.\frac{{\left( {\left( {2x + 1} \right)\ln 3} \right)'}}{{{{\left( {\left( {2x + 1} \right)\ln 3} \right)}^2}}}\\ =  - 2\frac{{2\ln 3}}{{{{\left( {\left( {2x + 1} \right)\ln 3} \right)}^2}}} = \frac{{ - 4\ln 3}}{{{{\left( {\left( {2x + 1} \right)\ln 3} \right)}^2}}}\end{array}\)

Tại \({x_0} = 3 \Rightarrow y''(3) = \frac{{ - 4\ln 3}}{{{{\left( {\left( {2.3 + 1} \right)\ln 3} \right)}^2}}} = \frac{{ - 4\ln 3}}{{{{\left( {7\ln 3} \right)}^2}}} = \frac{{ - 4}}{{49\ln 3}}\)

c, \(y' = 4{e^{4x + 3}} \Rightarrow y'' = 16{e^{4x + 3}}\)

Tại \({x_0} = 1 \Rightarrow y''(1) = 16.{e^{4.1 + 3}} = 16.{e^7}\)

d,

\(y' = 2\cos \left( {2x + \frac{\pi }{3}} \right) \Rightarrow y'' =  - 4\sin \left( {2x + \frac{\pi }{3}} \right)\)

Tại \({x_0} = \frac{\pi }{6} \Rightarrow y''\left( {\frac{\pi }{6}} \right) =  - 4\sin \left( {2.\frac{\pi }{6} + \frac{\pi }{3}} \right) =  - 2\sqrt 3 \)

e,

\(y' =  - 3.\sin \left( {3x - \frac{\pi }{6}} \right) \Rightarrow y'' =  - 9.\cos \left( {3x - \frac{\pi }{6}} \right)\)

Tại \({x_0} = 0 \Rightarrow y''(0) =  - 9.\cos \left( {3.0 - \frac{\pi }{6}} \right) = \frac{{ - 9\sqrt 3 }}{2}\)

tham khảo:

a)\(y'\left(x\right)=5\left(\dfrac{2x-1}{x+2}\right)^4.\dfrac{\left(x+2\right)\left(2\right)-\left(2x-1\right).1}{\left(x+2\right)^2}\)

\(=\dfrac{10\left(2x-1\right)\left(x+2\right)^3}{\left(x+2\right)^4}=\dfrac{20x-50}{\left(x+2\right)^4}\)

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

c)\(y'\left(x\right)=e^x.2sinxcosx+e^xsin^2x.2cosx\)

\(=2e^xsinx\left(cosx+sinxcosx\right)\)

\(=2e^xsinxcos^2x\)

d)\(y'\left(x\right)=\dfrac{1}{x\sqrt{x}}.\left(+\dfrac{1}{2\sqrt{x}}\right)\)

\(=\dfrac{1}{\sqrt{x}\left(2\sqrt{x}+\sqrt{x}+2\right)}\)

\(=\dfrac{1}{\sqrt{x}\left(3\sqrt{x}+2\right)}\)