"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.