How do you find the second derivative of # ln^2 (x)#?

1 Answer
Jul 11, 2016

#:. (d^2y)/(dx^2)={2(1-lnx)}/x^2=ln{(e/x)^(2/x^2)}.#

Explanation:

Let #y=ln^2(x)=(lnx)^2#.

#:. dy/dx=2lnx*d/dx(lnx)#.......[Chain Rule].....#=(2lnx)/x.#

#:. (d^2y)/(dx^2)=d/dx{dy/dx}#........................[Defn.]
#=d/dx{(2lnx)/x}#
#=2[{x*d/dx(lnx)-(lnx)*d/dx(x)}/x^2]#..........[Quotient Rule]
#=2[{x*1/x-(lnx)(1)}/x^2]#
#={2(1-lnx)}/x^2#
#=(2/x^2)(lne-lnx)#
#=(2/x^2)*ln(e/x)#
#=ln{(e/x)^(2/x^2)}#

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