tính đạo hàm cấp 2
a) \(y=2xe^x\)
b) \(y=3xlnx\)
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a: \(y'=\left(x^2-x\right)'=2x-1\)
\(y''=\left(2x-1\right)'=2\)
b: \(y'=\left(cosx\right)'=-sinx\)
\(y''=\left(-sinx\right)'=-cosx\)
\(a,y'=8x^3-10x\\ \Rightarrow y''=24x^2-10\\ b,y'=e^x+xe^x\\ \Rightarrow y''=e^x+e^x+xe^x=2e^x+xe^x\)
\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)
a: y=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: \(y=ln\left(2x+3\right)\)
=>\(y'=\dfrac{\left(2x+3\right)'}{\left(2x+3\right)}=\dfrac{2}{2x+3}\)
=>\(y''=\left(\dfrac{2}{2x+3}\right)'=\dfrac{2\left(2x+3\right)'-2'\left(2x+3\right)}{\left(2x+3\right)^2}\)
\(=\dfrac{4}{\left(2x+3\right)^2}\)
a: \(y=2\cdot x\cdot e^x\)
=>\(y'=2\cdot\left[x'\cdot e^x+\left(e^x\right)'\cdot x\right]\)
=>\(y'=2\cdot\left(e^x+e^x\cdot x\right)\)
=>\(y''=2\cdot\left[\left(e^x\right)'+\left(e^x\cdot x\right)'\right]\)
=>\(y''=2\cdot\left[e^x+\left(e^x\right)'\cdot x+e^x\cdot x'\right]\)
=>\(y''=2\cdot\left(e^x+e^x\cdot x+e^x\right)\)
=>\(y''=2\cdot e^x\left(x+2\right)\)
b: \(y=3\cdot x\cdot lnx\)
=>\(y'=3\left[x'\cdot lnx+x\cdot\left(lnx\right)'\right]\)
=>\(y'=3\left[lnx+x\cdot\dfrac{1}{x}\right]=3\cdot\left(lnx+1\right)\)
=>\(y''=3\left[\left(lnx\right)'+1'\right]=3\cdot\dfrac{1}{x}=\dfrac{3}{x}\)