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y=ln[x(2x+1)]=ln(2x^2+x)所以:y'=[1/(2x^2+x)]*(2x^2+x)'=[1/(2x^2+x)]*(4x+1)=(4x+1)/(2x^2+x).如果是:y=lnx*(2x+1),則:y=(1/x)(2x+1)+lnx*2=2+(1/x)+2lnx.=2(1+lnx)+(1/x).
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