\(y'\left(x\right)=\left(x^2\right)'sinx+x^2.\left(sinx\right)'\)\(=2x.sinx+x^2.cosx\)
\(y''\left(x\right)=\left[2x.sinx+x^2.cosx\right]'\)\(=\left(2xsinx\right)'\)\(+\left(x^2.cosx\right)'\)
\(=2sinx+2xcosx+2x.cosx+x^2.\left(-sinx\right)\)\(=2sinx+4x.cosx-x^2sinx\).
 

\(y'=\left(sin3x\right)'=3cos3x\)

\(\Rightarrow y''=-9sin3x\)

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

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

a,

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

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

Đáp án C

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