Answer
$$\eqalign{
& f'\left( x \right) = 3.2{x^{ - 1}} \cr
& f''\left( x \right) = - 3.2{x^{ - 2}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 3.2\ln x + 7.1 \cr
& \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left( {3.2\ln x} \right) + \frac{d}{{dx}}\left( {7.1} \right) \cr
& f'\left( x \right) = 3.2\frac{d}{{dx}}\left( {\ln x} \right) + \frac{d}{{dx}}\left( {7.1} \right) \cr
& {\text{compute derivatives}}{\text{, recall that 7}}{\text{.1 is a constant}}{\text{. then}}{\text{,}} \cr
& f'\left( x \right) = 3.2\left( {\frac{1}{x}} \right) \cr
& f'\left( x \right) = \frac{{3.2}}{x} \cr
& {\text{use }}\frac{1}{u} = {u^{ - 1}} \cr
& f'\left( x \right) = 3.2{x^{ - 1}} \cr
& \cr
& {\text{Differentiate to find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left( {3.2{x^{ - 1}}} \right) \cr
& {\text{use the power rule}} \cr
& f''\left( x \right) = 3.2\left( { - {x^{ - 2}}} \right) \cr
& f''\left( x \right) = - 3.2{x^{ - 2}} \cr} $$
