We could use the quotient rule to calculate the derivative, but instead we will break the function into a sum of parts and differentiate them individually. From part (d), we have
f ′ ( x ) = 2 − x x 3 = 2 x 3 − x x 3 = − 1 x 2 + 2 x 3 = − x − 2 + 2 x − 3 {\displaystyle f'(x)={\frac {2-x}{x^{3}}}={\frac {2}{x^{3}}}-{\frac {x}{x^{3}}}={\frac {-1}{x^{2}}}+{\frac {2}{x^{3}}}=-x^{-2}+2x^{-3}}
Then f ″ ( x ) = d d x ( − x − 2 ) + d d x ( 2 x − 3 ) . {\displaystyle f''(x)={\frac {d}{dx}}(-x^{-2})+{\frac {d}{dx}}(2x^{-3}).} And using the Power Rule of differentiation gives
d d x ( − x − 2 ) + d d x ( 2 x − 3 ) = 2 x − 3 − 6 x − 4 = 2 x 3 − 6 x 4 . {\displaystyle {\frac {d}{dx}}(-x^{-2})+{\frac {d}{dx}}(2x^{-3})=2x^{-3}-6x^{-4}={\frac {2}{x^{3}}}-{\frac {6}{x^{4}}}.}
Hence,
f ″ ( x ) = 2 x 3 − 6 x 4 . {\displaystyle \color {blue}f''(x)={\frac {2}{x^{3}}}-{\frac {6}{x^{4}}}.}
And using the Power Rule of differentiation gives
