How do you find the second derivative of $Y=(ln^2[x])/x$ ?

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Answer 1 · 2016-12-21T04:57:29

$y''=frac{2(-(lnx)^2-3lnx+1)}{x^3}$

$y=frac{(lnx)^2}{x}$

$y'=frac{(x)(2)(lnx)(1/x)-(lnx)^2(1)}{x^2}$

$y'=frac{2(lnx)-(lnx)^2}{x^2}$

$y'=frac{(lnx)(2-lnx)}{x^2}$

$y''=frac{(x^2)[(lnx)(-1/x)+(1/x)(2-lnx)]-(lnx)(2-lnx)(2x)}{(x^2)^2}$

$y''=frac{x(frac{-lnx+2-lnx}{x})-2(2lnx-(lnx)^2)}{x^3}$

$y''=frac{2-2lnx-4lnx-2(lnx)^2}{x^3}$

$y''=frac{2(-(lnx)^2-3lnx+1)}{x^3}$

How do you find the second derivative of Y=(ln^2[x])/xY=(ln^2[x])/x?