We can rewrite # f(x)=(lnx)^2/x^2 # as # f(x) = (lnx/x)^2 #
Now, let #u=lnx/x# Then using the product rule:
# d/dx(u/v)=( v(du)/dx-u(dv)/dx ) / v^2 #
So # (du)/dx = ( xd/dx(lnx)-(lnx)d/dx(x) ) / x^2 #
# :. (du)/dx = ( x1/x-(lnx)1 ) / x^2 #
# :. (du)/dx = ( 1-lnx ) / x^2 #
And using the chain rule we now have:
# f'(x) = d/dx{(lnx/x)^2 }#
# :. f'(x) = d/dxu^2 #
# :. f'(x) = d/(du)u^2 xx (du)/dx#
# :. f'(x) = 2u (du)/dx#
And substituting #u# and #(du)/dx# gives:
# :. f'(x) = 2lnx/x xx ( 1-lnx ) / x^2 #
# :. f'(x) = (2lnx( 1-lnx ) )/ x^3 #
