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\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)
\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)
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)\(y'=\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\)
\(y'=\dfrac{sin^2x-cos^2x}{sin^2x.cos^2x}\)
\(a,y'=\left(f\left(g\left(x\right)\right)\right)'\)
\(=f'\left(g\left(x\right)\right).g'\left(x\right)\)
\(=e^{g\left(x\right)}.\left(2x-1\right)\)
\(=e^{x^2-x}.\left(2x-1\right)\)
\(b,y'=\dfrac{d}{dx}\left(3^{sinx}\right)\)
\(=\dfrac{d}{dx}\left(e^{ln3.sinx}\right)\)
\(=\dfrac{d}{dx}\left(ln3.sinx\right).e^{ln3.sinx}\)
\(=ln3.cosx.3^{sinx}\)
a: \(y'=\left(sin3x\right)'+\left(sin^2x\right)'=3\cdot cos3x+sin\left(x+pi\right)\)
b: \(y'=\left(log_2\left(2x+1\right)\right)'+\left(3^{-2x+1}\right)'\)
\(=\dfrac{2}{\left(2n+1\right)\cdot ln2}-2\cdot3^{-2x+1}\cdot ln3\)
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 \({x_0} = 0 \Rightarrow y''(0) = - 9.\cos \left( {3.0 - \frac{\pi }{6}} \right) = \frac{{ - 9\sqrt 3 }}{2}\)
y ' = 2 2 − 3 x 2 x + 1 . 2 − 3 x 2 x + 1 ' = 2 2 − 3 x 2 x + 1 . − 3 2 x + 1 − 2 2 − 3 x 2 x + 1 2 = 2 2 − 3 x 2 x + 1 . − 7 2 x + 1 2 = − 14 2 x + 1 2 . 2 − 3 x 2 x + 1
Chọn đáp án A