Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (g)/Solution 1

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

$f^{'} (x) = \frac{2 - x}{x^{3}} = \frac{2}{x^{3}} - \frac{x}{x^{3}} = \frac{- 1}{x^{2}} + \frac{2}{x^{3}} = - x^{- 2} + 2 x^{- 3}$

Then $f^{″} (x) = \frac{d}{d x} (- x^{- 2}) + \frac{d}{d x} (2 x^{- 3}) .$ And using the Power Rule of differentiation gives

$\frac{d}{d x} (- x^{- 2}) + \frac{d}{d x} (2 x^{- 3}) = 2 x^{- 3} - 6 x^{- 4} = \frac{2}{x^{3}} - \frac{6}{x^{4}} .$

Hence,

$f^{″} (x) = \frac{2}{x^{3}} - \frac{6}{x^{4}} .$