# "We can start by rewriting the function, using Rules of" #
# "Logarithms, to prepare it for differentiation:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ ln( x^3 ) #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ 3 ln( x ) \qquad \qquad \qquad \quad \ color{blue}{ "Power Rule for Logs" } #
# "Now differentiation is much easier !!" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ 3 [ \ ln( x ) \ ]' #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ 3 cdot 1/x #
# "Continue, writing" \ f'(x) \ "to prepare it for differentiation:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f'(x) \ = \ 3 x^{-1}. #
# "Now differentiation is much easier -- no Quotient Rule needed !!" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f''(x) \ = \ 3 [ x^{-1} ]'. #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f''(x) \ = \ 3 [ (-1) x^{-2} ] \ = \ -3 x^{-2} . #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f''(x) \ = \ -3 cdot 1/x^2 \ = \ - 3/x^2 . #
# \qquad \qquad :. \qquad \qquad \qquad \qquad \quad \ f''(x) \ = \ - 3/x^2 . #
# "This is our answer." #
# "Summarizing:" #
# \qquad \qquad \qquad \qquad f(x) \ = \ ln( x^3 ) \qquad rArr \qquad f''(x) \ = \ - 3/x^2 quad. #